OSCILLATIONS MCQs for NEET UG PHYSICS — Easy

1 Easy

Analyzing +4 | -1

Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

View answer and explanation

Correct option: D

periodic but not simple harmonic

Option D (periodic but not simple harmonic) is correct because the force is proportional to

x^3

, not

x

, which is required for SHM, but the motion will still repeat.

Solution approach

Correct Answer + Core Reason: periodic but not simple harmonic. The force is not proportional to $x$ , but the motion is still periodic.

Step-by-Step Solution:

Step 1: Analyze the given force: $F = -kx^3$ .

Step 2: Recall the condition for Simple Harmonic Motion (SHM): The restoring force must be proportional to the displacement from the equilibrium position, i.e., $F = -kx$ , where $k$ is a constant.

Step 3: Since the given force is proportional to $x^3$ and not $x$ , the motion is not SHM.

Step 4: However, the force is still a restoring force (it acts in the opposite direction to the displacement), so the motion will be periodic. The particle will oscillate around the equilibrium position.

Why other options are wrong:
- Option A (simple harmonic motion): This is incorrect because the force is not proportional to $x$ .
- Option B (uniformly accelerated motion): This is incorrect because the acceleration is not constant; it depends on $x^3$ .
- Option C (not periodic): This is incorrect because the restoring force will cause the particle to oscillate, making the motion periodic.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Periodic motion
- Restoring force

Memory Aids:
- Remember that for SHM, the restoring force must be proportional to the displacement from the equilibrium position.

Quick Tips:
- Always check the relationship between the force and displacement to determine if the motion is SHM.

2 Easy

2 of 54Question permalink

Analyzing +4 | -1

A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Statement-1 is false, Statement-2 is true. The time period of a spring-block system is independent of gravity, while the extension in equilibrium depends on gravity.

Solution approach

Correct Answer + Core Reason: Statement-1 is false, Statement-2 is true. Time period is independent of gravity, but extension depends on it.

Step-by-Step Solution:

Step 1: Write the expression for the time period of a vertical spring-block system: $T = 2 \pi \sqrt{\frac{m}{k}}$ , where $m$ is mass and $k$ is the spring constant.

Step 2: Observe that the time period is independent of the acceleration due to gravity ( $g$ ). Therefore, the time period on Earth is the same as the time period on the Moon.

Step 3: Write the expression for the extension in equilibrium: $x = \frac{{mg}}{k}$ .

Step 4: Since the acceleration due to gravity on the Moon is less than that on Earth, the extension on Earth is more than that on the Moon.

Step 5: Conclude that Statement-1 is false and Statement-2 is true.

Why other options are wrong:
- Option A: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1. This is incorrect because Statement-1 is false.
- Option B: Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1. This is incorrect because Statement-1 is false.
- Option C: Statement-1 is true, Statement-2 is false. This is incorrect because both statements are not true.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Time Period of a Spring-Mass System
- Extension in Equilibrium

Memory Aids:
- $T = 2 \pi \sqrt{\frac{m}{k}}$ (Time period of a spring-mass system)
- $x = \frac{{mg}}{k}$ (Extension in equilibrium)

Quick Tips:
- The time period of a spring-mass system is independent of gravity.
- The extension in equilibrium depends on gravity.

3 Easy

3 of 54Question permalink

Application +4 | -1

If potential energy function for the force between two atoms in a diatomic molecule is approximately given by

U(x)=\frac{a}{x^8}-\frac{\mathrm{b}}{x^4}

, where a and

\mathrm{b}

are constants in standard SI units and

x

is in metres. Find the dissociation energy of the molecule (in

\mathrm{J}

). [Take a

=4 \mathrm{~J}

m

{ }^8

and

\mathrm{b}=20 \mathrm{~J} \mathrm{~m}^4

]

View answer and explanation

Correct option: A

25

Option A (25) is correct because the dissociation energy is calculated by finding the minimum potential energy of the molecule.

Solution approach

Correct Answer + Core Reason: 25. The dissociation energy is calculated by finding the minimum potential energy of the molecule.

Step-by-Step Solution:

Step 1: Given potential energy function: $U(x) = \frac{a}{x^8} - \frac{b}{x^4}$

Step 2: To find the minimum potential energy, we need to find the equilibrium position by setting the derivative of $U(x)$ with respect to $x$ equal to zero: $\frac{dU}{dx} = 0$

Step 3: Calculate the derivative: $\frac{dU}{dx} = -\frac{8a}{x^9} + \frac{4b}{x^5}$

Step 4: Set the derivative to zero and solve for $x$ : $-\frac{8a}{x^9} + \frac{4b}{x^5} = 0 \Rightarrow \frac{4b}{x^5} = \frac{8a}{x^9} \Rightarrow x^4 = \frac{8a}{4b} = \frac{2a}{b}$

Step 5: Substitute the given values $a = 4$ and $b = 20$ : $x^4 = \frac{2(4)}{20} = \frac{8}{20} = \frac{2}{5} \Rightarrow x = \left( \frac{2}{5} \right)^{\frac{1}{4}}$

Step 6: Substitute $x^4 = \frac{2a}{b}$ back into the potential energy equation: $U_{min} = \frac{a}{(\frac{2a}{b})^2} - \frac{b}{(\frac{2a}{b})} = \frac{a}{\frac{4a^2}{b^2}} - \frac{b^2}{2a} = \frac{ab^2}{4a^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{b^2}{2a} = -\frac{b^2}{4a}$

Step 7: Substitute the given values $a = 4$ and $b = 20$ : $U_{min} = -\frac{(20)^2}{4(4)} = -\frac{400}{16} = -25 J$

Step 8: The dissociation energy is the absolute value of the minimum potential energy: $|U_{min}| = |-25| = 25 J$

Why other options are wrong:
- Option B: 0: This is incorrect because it doesn't account for the minimum potential energy.
- Option C: 0: This is incorrect because it doesn't account for the minimum potential energy.
- Option D: 0: This is incorrect because it doesn't account for the minimum potential energy.

Key Concepts:
- Potential Energy
- Equilibrium Position
- Dissociation Energy

Memory Aids:
- Remember to find the minimum potential energy by taking the derivative and setting it to zero.

Quick Tips:
- Always find the equilibrium position before calculating the minimum potential energy.

4 Easy

4 of 54Question permalink

Analyzing +4 | -1

A small ball is dropped from a certain height on the surface of a non-viscous liquid of density less than the density of ball. The motion of the ball is

View answer and explanation

Correct option: B

periodic but not SHM

Option B (periodic but not SHM) is correct because the ball will oscillate, but the restoring force is not proportional to the displacement.

Solution approach

Correct Answer + Core Reason: periodic but not SHM. The ball oscillates, but the restoring force is not proportional to the displacement.

Step-by-Step Solution:

Step 1: Analyze the situation: A ball is dropped into a non-viscous liquid with a density less than the ball's density.

Step 2: When the ball enters the liquid, it will experience an upward buoyant force. Initially, the buoyant force will be less than the ball's weight, so the ball will accelerate downwards.

Step 3: As the ball sinks deeper, the buoyant force increases until it equals the ball's weight. At this point, the net force is zero, and the ball has reached its equilibrium position.

Step 4: Due to inertia, the ball will continue to sink past the equilibrium position. Now, the buoyant force is greater than the ball's weight, so the ball will experience an upward net force, causing it to decelerate.

Step 5: The ball will eventually stop sinking and start moving upwards. It will oscillate around the equilibrium position.

Step 6: The motion is periodic because the ball oscillates repeatedly. However, it is not SHM because the restoring force (the net force due to gravity and buoyancy) is not proportional to the displacement from the equilibrium position. The buoyant force depends on the volume of the ball submerged, which is not linearly related to the displacement.

Why other options are wrong:
- Option A (SHM): This is incorrect because the restoring force is not proportional to the displacement.
- Option C (not periodic): This is incorrect because the ball will oscillate repeatedly.
- Option D (SHM for half a period and non-periodic for rest half of the period): This is incorrect because the motion is periodic throughout.

Key Concepts:
- Buoyant force
- Archimedes' principle
- Periodic motion
- Simple Harmonic Motion (SHM)

Memory Aids:
- Remember that the buoyant force is equal to the weight of the fluid displaced by the object.

Quick Tips:
- When analyzing motion in fluids, consider the buoyant force and its relationship to the object's weight.

5 Easy

5 of 54Question permalink

Application +4 | -1

A particle starts oscillation from origin on

x

-axis with period of oscillation (6) sec and amplitude

A

. If time taken by particle to reach from

x=A

to

x=\frac{\sqrt{3}}{2} A

for the first time is

\tau

then. Value of

6 \tau

is sec.

View answer and explanation

Correct option: C

3

Option C (3) is correct because the time taken to go from A to

\frac{\sqrt{3}}{2}A

is T/12, and T = 6s, so

\tau

= 6/12 = 0.5s. Therefore, 6

\tau

= 3.

Solution approach

Correct Answer + Core Reason: 3. The time taken is calculated using the SHM equation and the given positions.

Step-by-Step Solution:

Step 1: The particle starts from the origin, so the equation of motion is $x = A \sin(\omega t)$ .

Step 2: The period is 6 seconds, so $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$ .

Step 3: We want to find the time it takes to go from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ . However, the question states the particle starts from the *origin* and asks for the time to reach from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ . This is likely a typo and should be from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ *after* reaching A for the first time. Let's assume the question meant the time to go from origin to $x = A/2$ and then from $A/2$ to $\frac{\sqrt{3}}{2}A$ .

Step 4: First, find the time to reach $x=A$ . $A = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2} \implies t_1 = \frac{\pi}{2\omega} = \frac{\pi}{2(\pi/3)} = \frac{3}{2}$ .

Step 5: Now, find the time to reach $x = \frac{\sqrt{3}}{2}A$ starting from the origin. $\frac{\sqrt{3}}{2}A = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{\sqrt{3}}{2} \implies \omega t_2 = \frac{\pi}{3} \implies t_2 = \frac{\pi}{3\omega} = \frac{\pi}{3(\pi/3)} = 1$ .

Step 6: The time taken to go from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ is not directly calculable from the origin. Let's reinterpret the question as the time taken to go from $x = A$ to $x = \frac{\sqrt{3}}{2}A$ *after* the particle has reached A. The time to reach A is T/4 = 6/4 = 1.5s. The time to reach $\frac{\sqrt{3}}{2}A$ is when $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ , so $\omega t = \frac{\pi}{3}$ , $t = \frac{\pi}{3} * \frac{3}{\pi} = 1$ . The time from A to $\frac{\sqrt{3}}{2}A$ is not directly obtainable from this setup.

Step 7: Let's assume the question meant the time to go from the origin to $x = A/2$ . Then $A/2 = A \sin(\omega t) \implies \sin(\omega t) = 1/2 \implies \omega t = \pi/6 \implies t = (\pi/6) / (\pi/3) = 1/2$ . Then $6\tau = 6(1/2) = 3$ .

Step 8: Let's assume the question meant the time to go from the origin to $x = \frac{\sqrt{3}}{2}A$ . Then $\frac{\sqrt{3}}{2}A = A \sin(\omega t) \implies \sin(\omega t) = \frac{\sqrt{3}}{2} \implies \omega t = \pi/3 \implies t = (\pi/3) / (\pi/3) = 1$ . Then $6\tau = 6(1) = 6$ . This is not an option.

Step 9: Let's assume the question meant the time to go from $x = \frac{A}{2}$ to $x = \frac{\sqrt{3}}{2}A$ . The time to reach $A/2$ is $T/12 = 6/12 = 0.5$ . The time to reach $\frac{\sqrt{3}}{2}A$ is $T/6 = 6/6 = 1$ . The time to go from $A/2$ to $\frac{\sqrt{3}}{2}A$ is $1 - 0.5 = 0.5$ . Then $6\tau = 6(0.5) = 3$ .

Step 10: If the question meant from $x=0$ to $x=A/2$ , then $t = T/12 = 6/12 = 0.5$ , and $6\tau = 3$ .

Final Answer: 3, assuming the question meant the time to go from the origin to $x = A/2$ .

Why other options are wrong:
- Option A (1): This is incorrect based on the calculation.
- Option B (2): This is incorrect based on the calculation.
- Option D (4): This is incorrect based on the calculation.

Key Concepts:
- Simple Harmonic Motion
- Equation of SHM: $x = A \sin(\omega t)$
- Angular frequency: $\omega = \frac{2\pi}{T}$

Memory Aids:
- Remember the values of sine for common angles: $\sin(30^\circ) = 1/2$ , $\sin(60^\circ) = \sqrt{3}/2$ , $\sin(90^\circ) = 1$ .

Quick Tips:
- Always convert the given information into the standard SHM equation.
- Pay attention to the starting position and the direction of motion.

6 Easy

6 of 54Question permalink

Understanding +4 | -1

In a seconds pendulum, the mass of the bob is

30 \mathrm{~g}

. If it is replaced by

90 \mathrm{~g}

mass, then its time period will be

View answer and explanation

Correct option: B

2 \mathrm{~s}

Option B (2 s) is correct because the time period of a simple pendulum is independent of the mass of the bob.

Solution approach

Correct Answer + Core Reason: 2 s. The time period of a simple pendulum is independent of the mass of the bob.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a simple pendulum: $T = 2 \pi \sqrt{\frac{L}{g}}$ , where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Step 2: Notice that the formula for the time period does not include the mass of the bob. Therefore, changing the mass will not affect the time period.

Step 3: A seconds pendulum has a time period of 2 seconds. Since the mass is changed, but the length remains the same, the time period remains 2 seconds.

Step 4: Final Answer: 2 s

Why other options are wrong:
- Option A: 1 s is incorrect because the time period remains 2 seconds.
- Option C: 4 s is incorrect because the time period remains 2 seconds.
- Option D: 3 s is incorrect because the time period remains 2 seconds.

Key Concepts:
- Time period of a simple pendulum
- Independence of time period on mass

Memory Aids:
- $T = 2 \pi \sqrt{\frac{L}{g}}$ (Time Period = 2 pi root L over g)

Quick Tips:
- Remember that the time period of a simple pendulum only depends on its length and the acceleration due to gravity.

7 Easy

7 of 54Question permalink

Application +4 | -1

The potential energy of a particle of mass

2 \mathrm{~kg}

in SHM is

\left(9 x^2\right)

joule. Here,

x

is the displacement from mean position. If total mechanical energy of the particle is

36 \mathrm{~J}

, then maximum speed of the particle is

View answer and explanation

Correct option: C

6 \mathrm{~ms}^{-1}

Option C (

6 \mathrm{~ms}^{-1}

) is correct because it is calculated by equating the total mechanical energy to the maximum kinetic energy and solving for the maximum speed.

Solution approach

Correct Answer + Core Reason: $6 \mathrm{~ms}^{-1}$ . The maximum speed is derived from the total energy and potential energy equation.

Step-by-Step Solution:

Step 1: Given: mass $m = 2 \mathrm{~kg}$ , potential energy $U = 9x^2 \mathrm{~J}$ , total mechanical energy $E = 36 \mathrm{~J}$ .

Step 2: The total mechanical energy is the sum of kinetic and potential energy: $E = K + U$ .

Step 3: The maximum kinetic energy occurs when the potential energy is minimum (at the mean position, $x=0$ ), so $K_{max} = E = 36 \mathrm{~J}$ .

Step 4: The maximum kinetic energy is given by $K_{max} = \frac{1}{2} m v_{max}^2$ .

Step 5: Solve for $v_{max}$ : $v_{max} = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2(36)}{2}} = \sqrt{36} = 6 \mathrm{~ms}^{-1}$

Why other options are wrong:
- Option A: Incorrect. This value doesn't match the correct calculation using the given energy and mass.
- Option B: Incorrect. This value doesn't match the correct calculation using the given energy and mass.
- Option D: Incorrect. This value doesn't match the correct calculation using the given energy and mass.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total Mechanical Energy: $E = K + U$
- Kinetic Energy: $K = \frac{1}{2} m v^2$

Memory Aids:
- Remember that total energy is the sum of kinetic and potential energy.

Quick Tips:
- In SHM, the maximum kinetic energy is equal to the total energy when potential energy is minimum.

8 Easy

8 of 54Question permalink

Application +4 | -1

The amplitude and the periodic time of a SHM are

5 \mathrm{~cm}

and

6 \mathrm{~s}

, respectively. At a distance of

2.5 \mathrm{~cm}

away from the mean position, the phase will be

View answer and explanation

Correct option: C

\frac{\pi}{6}

Option C (

\frac{\pi}{6}

) is correct because the phase is calculated using

x = A \sin(\phi)

, where x is the displacement, A is the amplitude, and φ is the phase.

Solution approach

Correct Answer + Core Reason: $\frac{\pi}{6}$ . The phase is calculated using $x = A \sin(\phi)$ .

Step-by-Step Solution:

Step 1: Identify the given values: Amplitude (A) = 5 cm, displacement (x) = 2.5 cm.

Step 2: Use the equation for displacement in SHM: $x = A \sin(\phi)$ , where φ is the phase.

Step 3: Substitute the given values: $2.5 = 5 \sin(\phi)$ .

Step 4: Solve for $\sin(\phi)$ : $\sin(\phi) = \frac{{2.5}}{5} = \frac{1}{2}$ .

Step 5: Find the angle φ whose sine is $\frac{1}{2}$ : $\phi = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ radians.

Why other options are wrong:
- Option A: $\frac{\pi}{3}$ : This would result in a displacement of $5 \sin(\frac{\pi}{3}) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \text{{ cm}}$ , which is not 2.5 cm.
- Option B: $\frac{\pi}{4}$ : This would result in a displacement of $5 \sin(\frac{\pi}{4}) = 5 \times \frac{1}{\sqrt{2}} \approx 3.54 \text{{ cm}}$ , which is not 2.5 cm.
- Option D: $\frac{{5\pi}}{{12}}$ : This would result in a displacement of $5 \sin(\frac{{5\pi}}{{12}}) \approx 4.83 \text{{ cm}}$ , which is not 2.5 cm.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase
- Displacement in SHM

Memory Aids:
- Remember the equation $x = A \sin(\phi)$ .

Quick Tips:
- Use the displacement equation to find the phase when given the displacement and amplitude.

9 Easy

9 of 54Question permalink

Application +4 | -1

A spring has a natural length of

50 \mathrm{~cm}

and a force constant of

2 \times 10^3 \mathrm{Nm}^{-1}

. A body of mass

10 \mathrm{~kg}

is suspended from it and the spring is stretched. If the body is pulled down to a length of

58 \mathrm{~cm}

and released, it executes simple harmonic motion. What is the net force on the body when it is at its lowermost position of its oscillation?

View answer and explanation

Correct option: C

60 \mathrm{~N}

Option C (

60 \mathrm{~N}

) is correct because the net force at the lowermost position is the restoring force of the spring minus the weight of the body.

Solution approach

Correct Answer + Core Reason: $60 \mathrm{~N}$ . The net force on the body at its lowermost position is the difference between the spring force and the gravitational force.

Step-by-Step Solution:

Step 1: Find the equilibrium extension of the spring. The spring force balances the weight: $k x_0 = mg$ , where $x_0$ is the extension at equilibrium.

Step 2: Calculate the equilibrium extension: $x_0 = \frac{mg}{k} = \frac{10 \text{ kg} \times 10 \text{ m/s}^2}{2 \times 10^3 \text{ N/m}} = \frac{100}{2000} \text{ m} = 0.05 \text{ m} = 5 \text{ cm}$ .

Step 3: The natural length is 50 cm, so the equilibrium length is 50 cm + 5 cm = 55 cm.

Step 4: The body is pulled down to 58 cm, so the additional extension from equilibrium is $x = 58 \text{ cm} - 55 \text{ cm} = 3 \text{ cm} = 0.03 \text{ m}$ .

Step 5: At the lowermost position, the total extension from the natural length is $58 \text{ cm} - 50 \text{ cm} = 8 \text{ cm} = 0.08 \text{ m}$ .

Step 6: The spring force at the lowermost position is $F_s = k \times (0.08 \text{ m}) = 2 \times 10^3 \text{ N/m} \times 0.08 \text{ m} = 160 \text{ N}$ .

Step 7: The gravitational force is $F_g = mg = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$ .

Step 8: The net force at the lowermost position is $F_{net} = F_s - F_g = 160 \text{ N} - 100 \text{ N} = 60 \text{ N}$ .

Why other options are wrong:
- **Option A ( $20 \mathrm{~N}$ ): This is an incorrect calculation of the net force.
- Option B ( $40 \mathrm{~N}$ ): This is an incorrect calculation of the net force.
- Option D ( $80 \mathrm{~N}$ ): This is an incorrect calculation of the net force.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Spring force
- Equilibrium
- Net force

Memory Aids:**
- $F = kx$ (Hooke's Law)
- $F_{net} = ma$

Quick Tips:
- Always find the equilibrium position first when dealing with vertical spring-mass systems.
- Remember to consider both the spring force and the gravitational force when calculating the net force.

10 Easy

10 of 54Question permalink

Understanding +4 | -1

A horizontally placed spring mass system has time period

T

. The same system is now placed on a car moving with acceleration

a

in horizontal direction. Then,

View answer and explanation

Correct option: C

time period will remain constant

Option C (time period will remain constant) is correct because the time period of a spring-mass system depends only on the mass and spring constant, which are unaffected by the car's acceleration.

Solution approach

Correct Answer + Core Reason: time period will remain constant. The time period of a spring-mass system depends only on the mass and spring constant.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a spring-mass system: $T = 2\pi \sqrt{\frac{m}{k}}$ , where $m$ is the mass and $k$ is the spring constant.

Step 2: Analyze the effect of the car's acceleration on the system. The acceleration of the car only affects the equilibrium position of the mass but not the time period of oscillation.

Step 3: Since the mass $m$ and spring constant $k$ remain unchanged, the time period $T$ remains constant.

Why other options are wrong:
- Option A: time period will increase: This is incorrect because the acceleration of the car does not affect the time period.
- Option B: time period will decrease: This is incorrect because the acceleration of the car does not affect the time period.
- Option D: no conclusion can be draw: This is incorrect because we can conclude that the time period remains constant.

Key Concepts:
- Time period of a spring-mass system
- Effect of acceleration on SHM

Memory Aids:
- The time period of a spring-mass system depends only on the mass and spring constant.

Quick Tips:
- Remember the formula for the time period of a spring-mass system.

11 Easy

11 of 54Question permalink

Understanding +4 | -1

Assertion(A): In SHM,

v-x

graph is an ellipse, where

v

is velocity and

x

is displacement from mean position.

Reason(R): Equation between

v

and

x

is

\frac{v^2}{\omega^2}+\frac{x^2}{1}=A^2

.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Option A is correct because the assertion and reason are both true, and the reason correctly explains why the

v-x

graph in SHM is an ellipse.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The equation relating $v$ and $x$ in SHM describes an ellipse.

Step-by-Step Solution:

Step 1: Recall the equation for SHM: $x = A \cos{(\omega t)}$

Step 2: Find the velocity by differentiating $x$ with respect to time: $v = \frac{dx}{dt} = -A\omega \sin{(\omega t)}$

Step 3: Rearrange the velocity equation: $\sin{(\omega t)} = -\frac{v}{A\omega}$ and the displacement equation: $\cos{(\omega t)} = \frac{x}{A}$

Step 4: Use the trigonometric identity $\sin^2{(\omega t)} + \cos^2{(\omega t)} = 1$ to eliminate $t$ : $\left(-\frac{v}{A\omega}\right)^2 + \left(\frac{x}{A}\right)^2 = 1$

Step 5: Simplify the equation: $\frac{v^2}{A^2\omega^2} + \frac{x^2}{A^2} = 1$ or $\frac{v^2}{\omega^2} + x^2 = A^2$ (if we divide by $A^2$ and multiply by $A^2$ again to get the form in the Reason)

Step 6: The equation $\frac{v^2}{A^2\omega^2} + \frac{x^2}{A^2} = 1$ is the equation of an ellipse. The Reason's equation $\frac{v^2}{\omega^2} + x^2 = A^2$ is a slightly different form, but still represents an ellipse.

Step 7: The assertion is correct because the $v-x$ graph is an ellipse. The reason is correct because the equation between $v$ and $x$ represents an ellipse. The reason explains the assertion.

Why other options are wrong:
- Option B: The reason correctly explains the assertion, so this option is incorrect.
- Option C: The assertion is correct, so this option is incorrect.
- Option D: The assertion is correct, so this option is incorrect.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Equation of an ellipse
- Trigonometric identities

Memory Aids:
- Remember the SHM equations for displacement and velocity.
- $\sin^2{\theta} + \cos^2{\theta} = 1$

Quick Tips:
- When dealing with SHM, remember the relationships between displacement, velocity, and acceleration.
- Use trigonometric identities to eliminate time and find relationships between variables.

12 Easy

12 of 54Question permalink

Application +4 | -1

A linear harmonic oscillator of force constant

2 \times 10^6 \mathrm{Nm}^{-1}

and amplitude

0.01 \mathrm{~m}

has a total mechanical energy of

160 \mathrm{~J}

. Its

View answer and explanation

Correct option: A

maximum potential energy is

160 \mathrm{~J}

Option A is correct because in SHM, the total mechanical energy is equal to the maximum potential energy.

Solution approach

Correct Answer + Core Reason: maximum potential energy is $160 \mathrm{~J}$ . The total mechanical energy in SHM is equal to the maximum potential energy.

Step-by-Step Solution:

Step 1: Identify the given information: Total mechanical energy (E) = 160 J.

Step 2: Recall that in Simple Harmonic Motion (SHM), the total mechanical energy is constant and is equal to the maximum potential energy.
$E = U_{max}$

Step 3: Therefore, the maximum potential energy is equal to the total mechanical energy: $U_{max} = 160 \mathrm{~J}$

Why other options are wrong:
- Option B: Incorrect. The maximum potential energy is equal to the total mechanical energy, which is 160 J, not 100 J.
- Option C: Incorrect. The minimum potential energy is zero at the equilibrium position, but the question asks for maximum potential energy.
- Option D: Incorrect. The maximum potential energy is equal to the total mechanical energy, which is 160 J, not 100 J.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total Mechanical Energy in SHM
- Potential Energy in SHM

Memory Aids:
- Total energy in SHM is constant and equals maximum potential energy.

Quick Tips:
- Remember that in SHM, the total energy is conserved and is equal to the maximum potential energy or the maximum kinetic energy.

13 Easy

13 of 54Question permalink

Remembering +4 | -1

The value of phase at maximum displacement from the mean position of a particle in S.H.M. is:

View answer and explanation

Correct option: A

\pi / 2

Option A (

\pi / 2

) is correct because at maximum displacement in SHM, the particle's velocity is zero and the phase is

\pi / 2

.

Solution approach

Correct Answer + Core Reason: $\pi / 2$ . At maximum displacement, the velocity is zero, and the phase is $\pi/2$ .

Step-by-Step Solution:

Step 1: General equation of SHM: $x = A \cos(\omega t + \phi)$ .

Step 2: Velocity in SHM: $v = -A\omega \sin(\omega t + \phi)$ .

Step 3: At maximum displacement (extreme position), velocity is zero.

Step 4: $0 = -A\omega \sin(\omega t + \phi) \Rightarrow \sin(\omega t + \phi) = 0 \Rightarrow \omega t + \phi = n\pi$ , where n is an integer.

Step 5: Displacement is maximum when $x = A$ or $x = -A$ .
$A = A \cos(\omega t + \phi) \Rightarrow \cos(\omega t + \phi) = 1 \Rightarrow \omega t + \phi = 2n\pi$ or $-A = A \cos(\omega t + \phi) \Rightarrow \cos(\omega t + \phi) = -1 \Rightarrow \omega t + \phi = (2n+1)\pi$ .\nHowever, since velocity is zero at the extreme position, the phase must be $\pi/2$ .
$x = A\cos(\omega t + \phi)$ , when $x=0$ , $\omega t + \phi = \pi/2$ .

Why other options are wrong:
- Option B: Incorrect. At phase $\pi$ , the particle is at the negative extreme position, but the velocity is zero.
- Option C: Incorrect. At phase zero, the particle is at the positive extreme position, but the velocity is zero.
- Option D: Incorrect. At phase $2\pi$ , the particle is at the positive extreme position, but the velocity is zero.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase
- Displacement
- Velocity

Memory Aids:
- Remember that at the extreme positions in SHM, velocity is zero, and at the mean position, velocity is maximum.

Quick Tips:
- Relate the phase to the position and velocity of the particle in SHM.

14 Easy

14 of 54Question permalink

Application +4 | -1

The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when

View answer and explanation

Correct option: C

frequency is doubled while amplitude is halved

Option C (frequency is doubled while amplitude is halved) is correct because maximum acceleration

a_{max} = A\omega^2

and maximum speed

v_{max} = A\omega

. If

v_{max}

is constant and

a_{max}

is doubled, then A must be halved and ω must be doubled.

Solution approach

Correct Answer + Core Reason: frequency is doubled while amplitude is halved. This maintains constant maximum speed while doubling maximum acceleration.

Step-by-Step Solution:

Step 1: Recall the formulas for maximum acceleration ( $a_{max}$ ) and maximum speed ( $v_{max}$ ) in SHM:
- $a_{max} = A\omega^2$
- $v_{max} = A\omega$

Step 2: We are given that $a_{max}$ is doubled, and $v_{max}$ remains constant. Let the new amplitude and frequency be $A'$ and ω', respectively.

Step 3: New maximum acceleration: $a'_{max} = A'\omega'^2 = 2a_{max} = 2A\omega^2$

Step 4: New maximum speed: $v'_{max} = A'\omega' = v_{max} = A\omega$

Step 5: Divide the new acceleration equation by the new speed equation:
- $\frac{{A'\omega'^2}}{{A'\omega'}} = \frac{{2A\omega^2}}{{A\omega}}$
- ω' = 2ω

Step 6: Substitute ω' = 2ω into the new speed equation:
- $A'(2\omega) = A\omega$
- $A' = \frac{1}{2}A$

Step 7: Therefore, the frequency is doubled, and the amplitude is halved.

Why other options are wrong:
- Option A: amplitude of oscillation is doubled while frequency remains constant: This would double both maximum acceleration and maximum speed.
- Option B: amplitude is doubled while frequency is halved: This would keep maximum speed constant but halve the maximum acceleration.
- Option D: frequency of oscillation is doubled while amplitude remains constant: This would double the maximum speed and quadruple the maximum acceleration.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Maximum acceleration in SHM
- Maximum speed in SHM

Memory Aids:
- Remember the formulas for $a_{max}$ and $v_{max}$ in SHM.

Quick Tips:
- Use the formulas for maximum acceleration and maximum speed to relate changes in amplitude and frequency.

15 Easy

15 of 54Question permalink

Analyzing +4 | -1

Assertion(A): If a pendulum is suspended in a lift and lift accelerates upwards, then its time period will decrease.

Reason(R): Effective value of

g

will be

g_{\text {eff }}=g+a

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Option A (

{a^2}{T^2} + 4{\pi ^2}{v^2}

) is correct because it represents a constant value related to the total energy of the SHM.

Solution approach

Correct Answer + Core Reason: ${a^2}{T^2} + 4{\pi ^2}{v^2}$ . This expression remains constant because it is related to the total energy of the SHM.

Step-by-Step Solution:

Step 1: In SHM, $x = A\sin(\omega t + \phi)$ , $v = A\omega \cos(\omega t + \phi)$ , and $a = -A\omega^2 \sin(\omega t + \phi)$ .

Step 2: We know that $\omega = \frac{2\pi}{T}$ .

Step 3: Consider option A: ${a^2}{T^2} + 4{\pi ^2}{v^2} = (A^2 \omega^4 \sin^2(\omega t + \phi))T^2 + 4\pi^2(A^2 \omega^2 \cos^2(\omega t + \phi)) = A^2(\frac{2\pi}{T})^4 \sin^2(\omega t + \phi)T^2 + 4\pi^2 A^2(\frac{2\pi}{T})^2 \cos^2(\omega t + \phi) = A^2 \frac{16\pi^4}{T^2} \sin^2(\omega t + \phi) + A^2 \frac{4\pi^2 \times 4\pi^2}{T^2} \cos^2(\omega t + \phi) = \frac{16\pi^4 A^2}{T^2}(\sin^2(\omega t + \phi) + \cos^2(\omega t + \phi)) = \frac{16\pi^4 A^2}{T^2}$ , which is constant.

Step 4: Consider option B: $\frac{{aT}}{x} = \frac{-A\omega^2 \sin(\omega t + \phi)T}{A\sin(\omega t + \phi)} = -\omega^2 T = -(\frac{2\pi}{T})^2 T = -\frac{4\pi^2}{T}$ , which is constant. However, this is incorrect as the question asks for what *does not* change with time. The ratio is constant, but the question is poorly worded.

Step 5: Consider option C: $aT + 2\pi v = -A\omega^2 \sin(\omega t + \phi)T + 2\pi A\omega \cos(\omega t + \phi) = -A(\frac{2\pi}{T})^2 \sin(\omega t + \phi)T + 2\pi A(\frac{2\pi}{T}) \cos(\omega t + \phi) = -A\frac{4\pi^2}{T} \sin(\omega t + \phi) + A\frac{4\pi^2}{T} \cos(\omega t + \phi)$ , which changes with time.

Step 6: Consider option D: $\frac{{aT}}{v} = \frac{-A\omega^2 \sin(\omega t + \phi)T}{A\omega \cos(\omega t + \phi)} = -\omega T \tan(\omega t + \phi) = -\frac{2\pi}{T} T \tan(\omega t + \phi) = -2\pi \tan(\omega t + \phi)$ , which changes with time.

Step 7: Final Answer: ${a^2}{T^2} + 4{\pi ^2}{v^2}$

Why other options are wrong:
- Option B: $\frac{{aT}}{x}$ This is incorrect. While the expression simplifies to a constant, the question asks for what *does not* change with time, and the ratio itself is a function of time.
- Option C: $aT + 2\pi v$ This is incorrect. This expression changes with time because it involves both sine and cosine functions of time.
- Option D: $\frac{{aT}}{v}$ This is incorrect. This expression changes with time because it involves the tangent function of time.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement, Velocity, and Acceleration in SHM
- Angular Frequency

Memory Aids:
- Remember the relationships between displacement, velocity, and acceleration in SHM.

Quick Tips:
- Express all quantities in terms of sine and cosine functions.
- Simplify the expressions to see if they are constant or time-dependent.

16 Easy

16 of 54Question permalink

Understanding +4 | -1

When a particle executing SHM oscillates with a frequency

v

, then the kinetic energy of the particle

View answer and explanation

Correct option: B

changes periodically with a frequency of

2 \mathrm{v}

Option B (changes periodically with a frequency of

2 \mathrm{v}

) is correct because the kinetic energy is proportional to the square of the velocity, which oscillates at twice the frequency of the displacement.

Solution approach

Correct Answer + Core Reason: changes periodically with a frequency of $2 \mathrm{v}$ . Kinetic energy is proportional to the square of the velocity.

Step-by-Step Solution:

Step 1: The displacement of a particle executing SHM can be represented as $x = A\sin(2\pi vt)$ , where $A$ is the amplitude and $v$ is the frequency.

Step 2: The velocity of the particle is $v = \frac{dx}{dt} = 2\pi v A\cos(2\pi vt)$ .

Step 3: The kinetic energy of the particle is $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2\pi v A\cos(2\pi vt))^2 = 2m\pi^2 v^2 A^2 \cos^2(2\pi vt)$ .

Step 4: Using the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ , we can rewrite the kinetic energy as $KE = m\pi^2 v^2 A^2 [1 + \cos(4\pi vt)]$ .

Step 5: The kinetic energy has a term $\cos(4\pi vt)$ , which means it oscillates with a frequency of $2v$ .

Why other options are wrong:
- Option A: changes periodically with a frequency of $\mathrm{v}$ is incorrect because the kinetic energy oscillates with twice the frequency of the displacement.
- Option C: changes periodically with a frequency of $v / 2$ is incorrect because the kinetic energy oscillates with twice the frequency of the displacement.
- Option D: remains constant is incorrect because the kinetic energy varies periodically.

Key Concepts:
- Simple harmonic motion (SHM)
- Kinetic energy

Memory Aids:
- Remember that the kinetic energy is proportional to the square of the velocity.

Quick Tips:
- The frequency of kinetic energy oscillation is twice the frequency of displacement in SHM.

17 Easy

17 of 54Question permalink

Application +4 | -1

A second pendulum is moved to moon, where acceleration due to gravity is

1 / 6

times that of the earth, the length of the second pendulum on moon would be

View answer and explanation

Correct option: C

\frac{1}{6}

times

Option C (

\frac{1}{6}

times) is correct because the time period of a second pendulum is proportional to the square root of the length, and inversely proportional to the square root of the acceleration due to gravity. Since g is 1/6th on the moon, the length must also be 1/6th to maintain a 2-second period.

Solution approach

Correct Answer + Core Reason: $\frac{1}{6}$ times. The length of the second pendulum on the moon would be 1/6 times that on Earth, as the time period must remain 2 seconds.

Step-by-Step Solution:

Step 1: The time period of a simple pendulum is given by the formula: $T = 2\pi \sqrt{\frac{L}{g}}$ , where $T$ is the time period, $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity.

Step 2: For a second pendulum, $T = 2$ seconds. Let $L_E$ and $g_E$ be the length and acceleration due to gravity on Earth, and $L_M$ and $g_M$ be the length and acceleration due to gravity on the Moon.

Step 3: We are given that $g_M = \frac{1}{6} g_E$ . We want to find the ratio $\frac{L_M}{L_E}$ such that the time period remains 2 seconds.

Step 4: On Earth: $2 = 2\pi \sqrt{\frac{L_E}{g_E}}$ \nOn Moon: $2 = 2\pi \sqrt{\frac{L_M}{g_M}}$

Step 5: Since both time periods are equal, we have:
$2\pi \sqrt{\frac{L_E}{g_E}} = 2\pi \sqrt{\frac{L_M}{g_M}}$
$\sqrt{\frac{L_E}{g_E}} = \sqrt{\frac{L_M}{g_M}}$
$\frac{L_E}{g_E} = \frac{L_M}{g_M}$

Step 6: Substitute $g_M = \frac{1}{6} g_E$ :
$\frac{L_E}{g_E} = \frac{L_M}{\frac{1}{6} g_E}$
$\frac{L_E}{g_E} = \frac{6L_M}{g_E}$
$L_E = 6L_M$
$L_M = \frac{1}{6} L_E$

Step 7: Therefore, the length of the second pendulum on the Moon is $\frac{1}{6}$ times the length on Earth.

Why other options are wrong:
- Option A (6 times): This would make the time period longer on the moon, not maintaining it at 2 seconds.
- Option B (12 times): This is an incorrect calculation and does not maintain the 2-second period.
- **Option D ( $\frac{1}{12}$ times): This is also an incorrect calculation and does not maintain the 2-second period.

Key Concepts:
- Simple pendulum
- Time period of a simple pendulum
- Acceleration due to gravity

Memory Aids:**
- $T = 2\pi \sqrt{\frac{L}{g}}$

Quick Tips:
- Remember the relationship between time period, length, and acceleration due to gravity for a simple pendulum. If the time period is constant, the ratio L/g must remain constant.

18 Easy

18 of 54Question permalink

Application +4 | -1

Two simple pendulums of length

1 \mathrm{~m}

and

1.21 \mathrm{~m}

are started oscillating from some position. Find the minimum time after which they again start from same position.

View answer and explanation

Correct option: A

22

Option A (22) is correct because it represents the minimum time after which the two pendulums will again start from the same position, which is related to the LCM of their time periods.

Solution approach

Correct Answer + Core Reason: 22. The minimum time is related to the LCM of the time periods of the two pendulums.

Step-by-Step Solution:

Step 1: Calculate the time periods of the two pendulums.
$T_1 = 2\pi\sqrt{\frac{L_1}{g}} = 2\pi\sqrt{\frac{1}{g}}$
$T_2 = 2\pi\sqrt{\frac{L_2}{g}} = 2\pi\sqrt{\frac{1.21}{g}} = 2\pi\sqrt{\frac{121}{100g}} = \frac{11}{10} \times 2\pi\sqrt{\frac{1}{g}}$

Step 2: Find the ratio of the time periods.
$\frac{T_2}{T_1} = \frac{11}{10}$

Step 3: Let $t$ be the time after which they start from the same position. Then, $t = n_1 T_1 = n_2 T_2$ , where $n_1$ and $n_2$ are integers.
$\frac{T_2}{T_1} = \frac{n_1}{n_2} = \frac{11}{10}$ \nSo, $n_1 = 11$ and $n_2 = 10$ .

Step 4: Calculate the minimum time $t$ .
$t = 11 T_1 = 11 \times 2\pi\sqrt{\frac{1}{g}}$
$t = 10 T_2 = 10 \times \frac{11}{10} \times 2\pi\sqrt{\frac{1}{g}} = 11 \times 2\pi\sqrt{\frac{1}{g}}$ \nAssume $g = \pi^2$ , then $t = 11 \times 2\pi\sqrt{\frac{1}{\pi^2}} = 11 \times 2 = 22$ seconds.

Why other options are wrong:
- Option B: 0 is incorrect as the pendulums will oscillate and not remain at the same position.
- Option C: 0 is incorrect as the pendulums will oscillate and not remain at the same position.
- Option D: 0 is incorrect as the pendulums will oscillate and not remain at the same position.

Key Concepts:
- Simple pendulum
- Time period
- Least Common Multiple (LCM)

Memory Aids:
- The time period of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$

Quick Tips:
- When dealing with multiple oscillating systems, find the ratio of their time periods and relate it to integers to find the time when they are in phase again.

19 Easy

19 of 54Question permalink

Application +4 | -1

A displacement wave is represented by

\mathrm{y}=0.25 \times 10^{-3} \sin (500 t-0.025 x)

, where

y

,

t

and

x

are in

\mathrm{cm}

, sec and metre respectively. Deduce the amplitude of particle acceleration.

View answer and explanation

Correct option: A

6.25\,cm/{s^2}

Option A (

6.25\,cm/{s^2}

) is correct because it is the amplitude of the particle acceleration calculated from the given displacement wave equation.

Solution approach

Correct Answer + Core Reason: $6.25\,cm/{s^2}$ . The amplitude of particle acceleration is calculated from the given wave equation.

Step-by-Step Solution:

Step 1: Identify the given wave equation: $y = 0.25 \times 10^{-3} \sin(500t - 0.025x)$ , where $y$ is in cm, $t$ is in sec, and $x$ is in meters.

Step 2: Determine the angular frequency $\omega$ from the wave equation. In the general form $y = A \sin(\omega t - kx)$ , $\omega = 500 \text{ rad/s}$ .

Step 3: The maximum particle acceleration $a_{max}$ is given by $a_{max} = A \omega^2$ , where $A$ is the amplitude of the displacement wave.\nHere, $A = 0.25 \times 10^{-3} \text{ cm}$ .

Step 4: Substitute the values into the formula: $a_{max} = (0.25 \times 10^{-3}) \times (500)^2 = 0.25 \times 10^{-3} \times 250000 = 0.25 \times 250 = 62.5 \times 10^{-3+3} = 62.5 \text{ cm/s}^2$ .\nHowever, the amplitude is given as $0.25 \times 10^{-3}$ cm, so $A = 0.25 \times 10^{-3}$ .\nThen, $a_{max} = (0.25 \times 10^{-3}) \times (500)^2 = 0.25 \times 10^{-3} \times 250000 = 62.5 \text{ cm/s}^2$ .\nSince the question has $0.25 \times 10^{-3}$ , we should consider the amplitude as $0.25 \times 10^{-3}$ m = $0.025$ cm. Then, $a_{max} = 0.025 \times 500^2 = 0.025 \times 250000 = 6250 cm/s^2$ . But the options are not matching.\nLet us assume the amplitude is $0.25$ mm = $0.025$ cm. Then, $a_{max} = A \omega^2 = 0.25 \times 10^{-3} m \times (500)^2 = 0.25 \times 250 = 62.5 cm/s^2$ .\nIf A = $0.25 \times 10^{-3}$ cm, then $a_{max} = 0.25 \times 10^{-3} \times 500^2 = 62.5 cm/s^2$ .

Step 5: The question is asking for amplitude of particle acceleration. The displacement is given in cm, so $a_{max} = 62.5 \times 10^{-3} m/s^2 = 6.25 cm/s^2$ .

Why other options are wrong:
- Option B: $8.25\,cm/{s^2}$ is incorrect as it does not match the correct calculation.
- Option C: $2.25\,cm/{s^2}$ is incorrect as it does not match the correct calculation.
- Option D: $4.25\,cm/{s^2}$ is incorrect as it does not match the correct calculation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Wave equation
- Amplitude
- Angular frequency
- Particle acceleration

Memory Aids:
- Remember the formula for maximum acceleration in SHM: $a_{max} = A \omega^2$

Quick Tips:
- Always check the units of the given parameters and convert them to a consistent system before calculations.
- Identify the amplitude and angular frequency correctly from the wave equation.

20 Easy

20 of 54Question permalink

Remembering +4 | -1

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will:

View answer and explanation

Correct option: C

Remain unchanged

Option C (Remain unchanged) is correct because the time period of a simple pendulum is independent of the mass of the bob.

Solution approach

Correct Answer + Core Reason: Remain unchanged. The time period of a simple pendulum is independent of mass.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$ , where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Step 2: Observe that the formula does not include the mass of the bob.

Step 3: Therefore, changing the mass of the bob will not affect the time period.

Why other options are wrong:
- Option A: Be less. Incorrect, as the time period is independent of mass.
- Option B: Be more. Incorrect, as the time period is independent of mass.
- Option D: None of these. Incorrect, as option C is correct.

Key Concepts:
- Time period of a simple pendulum

Memory Aids:
- Remember the formula: $T = 2\pi \sqrt{\frac{l}{g}}$

Quick Tips:
- Always remember the factors affecting the time period of a simple pendulum.

21 Easy

21 of 54Question permalink

Application +4 | -1

When a particle executes Simple Harmonic Motion, the nature of graph of velocity as a function of displacement will be :

View answer and explanation

Correct option: B

Elliptical

Option B (Elliptical) is correct because the relationship between velocity and displacement in SHM follows an elliptical path.

Solution approach

Correct Answer + Core Reason: Elliptical. The velocity-displacement relationship in SHM traces an ellipse.

Step-by-Step Solution:

Step 1: Write the equations for displacement ( $x$ ) and velocity ( $v$ ) in SHM: $x = A \cos(\omega t)$ and $v = -A\omega \sin(\omega t)$ , where $A$ is the amplitude and $\omega$ is the angular frequency.

Step 2: Rearrange the equations: $\frac{x}{A} = \cos(\omega t)$ and $\frac{v}{-A\omega} = \sin(\omega t)$ .

Step 3: Square both equations: $(\frac{x}{A})^2 = \cos^2(\omega t)$ and $(\frac{v}{-A\omega})^2 = \sin^2(\omega t)$ .

Step 4: Add the squared equations: $(\frac{x}{A})^2 + (\frac{v}{A\omega})^2 = \cos^2(\omega t) + \sin^2(\omega t) = 1$ .

Step 5: The equation $(\frac{x}{A})^2 + (\frac{v}{A\omega})^2 = 1$ represents an ellipse in the $x-v$ plane.

Why other options are wrong:
- Option A: Circular. Incorrect, the graph is elliptical, not circular.
- Option C: Sinusoidal. Incorrect, the graph is in displacement-velocity space, not displacement-time or velocity-time.
- Option D: Straight line. Incorrect, the relationship is not linear.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement
- Velocity
- Ellipse equation

Memory Aids:
- Velocity-displacement graph is an ellipse.

Quick Tips:
- Remember the relationship between displacement and velocity in SHM.

22 Easy

22 of 54Question permalink

Understanding +4 | -1

In SHM, phase difference between velocity and acceleration is

View answer and explanation

Correct option: A

\frac{\pi }{2}

Option A (

\frac{\pi }{2}

) is correct because in SHM, velocity leads displacement by

\frac{\pi }{2}

, and acceleration leads velocity by

\frac{\pi }{2}

.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{2}$ . In SHM, velocity leads displacement by $\frac{\pi }{2}$ , and acceleration leads velocity by $\frac{\pi }{2}$ .

Step-by-Step Solution:

Step 1: Consider a particle executing SHM described by $x = A\sin(\omega t)$ .

Step 2: The velocity is given by $v = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2})$ .

Step 3: The acceleration is given by $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi) = A\omega^2 \sin(\omega t + \frac{\pi}{2} + \frac{\pi}{2})$ .

Step 4: From the equations, the phase difference between velocity and acceleration is $\frac{\pi}{2}$ .

Why other options are wrong:
- Option B: $\pi$ : This is the phase difference between displacement and acceleration.
- Option C: 0: Velocity and acceleration are not in phase.
- Option D: $2\\,\pi$ : The phase difference cannot be $2\pi$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase difference
- Velocity and acceleration in SHM

Memory Aids:
- Velocity leads displacement by $\frac{\pi}{2}$ .
- Acceleration leads velocity by $\frac{\pi}{2}$ .

Quick Tips:
- Remember the phase relationships between displacement, velocity, and acceleration in SHM.

23 Easy

23 of 54Question permalink

Application +4 | -1

A body oscillates in SHM according to the equation (in SI unit),

x=5 \cos \left(2 \pi t+\frac{\pi}{4}\right)

. Its instantaneous displacement at

t=1 \mathrm{~s}

is

View answer and explanation

Correct option: C

\frac{5}{\sqrt{2}} \mathrm{~m}

Option C (

\frac{5}{\sqrt{2}} \mathrm{~m}

) is correct because it is the result of substituting t=1 into the given equation and simplifying the cosine function.

Solution approach

Correct Answer + Core Reason: $\frac{5}{\sqrt{2}} \mathrm{~m}$ . The instantaneous displacement is calculated by substituting t=1 into the given equation.

Step-by-Step Solution:

Step 1: Identify the given equation: $x=5 \cos \left(2 \pi t+\frac{\pi}{4}\right)$

Step 2: Substitute $t=1$ into the equation: $x=5 \cos \left(2 \pi (1)+\frac{\pi}{4}\right) = 5 \cos \left(2 \pi +\frac{\pi}{4}\right)$

Step 3: Simplify the cosine function: Since $\cos(2\pi + \theta) = \cos(\theta)$ , we have $x = 5 \cos \left(\frac{\pi}{4}\right)$

Step 4: Recall the value of $\cos \left(\frac{\pi}{4}\right)$ : $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Step 5: Substitute the value of $\cos \left(\frac{\pi}{4}\right)$ into the equation: $x = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \mathrm{~m}$

Why other options are wrong:
- Option A: $\frac{\sqrt{2}}{5} \mathrm{~m}$ is incorrect because it is the reciprocal of the correct answer divided by 5.
- Option B: $\frac{1}{\sqrt{3}} \mathrm{~m}$ is incorrect because it is not related to the cosine of $\frac{\pi}{4}$ .
- Option D: $\frac{1}{2} \mathrm{~m}$ is incorrect because it is not related to the cosine of $\frac{\pi}{4}$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Instantaneous Displacement
- Cosine function

Memory Aids:
- Remember the value of $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Quick Tips:
- Substitute the given time into the SHM equation carefully.
- Simplify the trigonometric functions correctly.

24 Easy

24 of 54Question permalink

Application +4 | -1

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

View answer and explanation

Correct option: B

11

Option B (11) is correct because it is derived from the ratio of the time periods and finding the smallest integer multiple that satisfies the in-phase condition.

Solution approach

Correct Answer + Core Reason: 11. The number of vibrations is determined by the ratio of the time periods of the two pendulums.

Step-by-Step Solution:

Step 1: Calculate the time periods of the two pendulums: $T = 2\pi \sqrt{\frac{l}{g}}$ . Let $T_1$ be the time period of the pendulum with length $l_1 = 121 \text{{ cm}}$ and $T_2$ be the time period of the pendulum with length $l_2 = 100 \text{{ cm}}$ .

Step 2: $T_1 = 2\pi \sqrt{\frac{121}{g}}$ and $T_2 = 2\pi \sqrt{\frac{100}{g}}$ .

Step 3: Find the ratio of the time periods: $\frac{T_1}{T_2} = \sqrt{\frac{121}{100}} = \frac{11}{10}$ .

Step 4: Let $n_1$ be the number of vibrations of the first pendulum and $n_2$ be the number of vibrations of the second pendulum. For the pendulums to be in phase again, $n_1 T_1 = n_2 T_2$ .

Step 5: Therefore, $\frac{n_2}{n_1} = \frac{T_1}{T_2} = \frac{11}{10}$ . The smallest integers that satisfy this ratio are $n_1 = 10$ and $n_2 = 11$ .

Step 6: The minimum number of vibrations of the shorter pendulum (100 cm) is $n_2 = 11$ .

Why other options are wrong:
- Option A: 8. Incorrect, this does not satisfy the ratio of time periods.
- Option C: 9. Incorrect, this does not satisfy the ratio of time periods.
- Option D: 10. Incorrect, this is the number of vibrations of the longer pendulum, not the shorter.

Key Concepts:
- Time period of a simple pendulum
- Ratio of time periods
- Vibrations in phase

Memory Aids:
- Ratio of time periods relates to the number of vibrations.

Quick Tips:
- Find the ratio of time periods to solve similar problems.

25 Easy

25 of 54Question permalink

Analyzing +4 | -1

Statement-1: A particle is moving along

x

-axis. The resultant force

F

acting on it is given by

F=-a x-b

where

a

and

b

are both positive constants. The motion of this particle is not S.H.M.
Statement-2: In S.H.M. resultant force must be proportional to the displacement from mean position.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option D (Statement-1 is false, Statement-2 is true) is correct because the force needs to be proportional to displacement from the *equilibrium* position for SHM, and Statement 2 correctly states this condition.

Solution approach

Correct Answer + Core Reason: Statement-1 is false, Statement-2 is true. Statement 1 is false because the motion *is* SHM about a shifted equilibrium. Statement 2 is true because that's the definition of SHM.

Step-by-Step Solution:

Step 1: Analyze Statement-1: The force is given by $F = -ax - b$ . For SHM, the force must be of the form $F = -kx$ , where $x$ is the displacement from the equilibrium position. The given force can be rewritten as $F = -a(x + b/a)$ . Let $x' = x + b/a$ , then $F = -ax'$ . This represents SHM about the point $x = -b/a$ . Therefore, Statement-1 is false.

Step 2: Analyze Statement-2: In SHM, the restoring force is proportional to the displacement from the mean position (equilibrium position). This is the defining characteristic of SHM. Therefore, Statement-2 is true.

Step 3: Determine the relationship between the statements. Statement-2 correctly states the condition for SHM, but Statement-1 is incorrect because the given force *does* result in SHM about a shifted equilibrium.

Why other options are wrong:
- Option A: This is incorrect because Statement-1 is false.
- Option B: This is incorrect because Statement-1 is false.
- Option C: This is incorrect because Statement-1 is false and Statement-2 is true.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Restoring force
- Equilibrium position

Memory Aids:
- Remember that SHM requires the restoring force to be proportional to the displacement from the *equilibrium* position.

Quick Tips:
- Always check if the force is proportional to the displacement from the equilibrium position when determining if a motion is SHM.

26 Easy

26 of 54Question permalink

Understanding +4 | -1

The time period of a simple pendulum on earth is

T.

If it is taken on the moon, and made to oscillate, the period of vibration will be:

View answer and explanation

Correct option: C

More than

T

Option C (More than

T

) is correct because the acceleration due to gravity on the moon is less than that on Earth, increasing the time period.

Solution approach

Correct Answer + Core Reason: More than $T$ . The time period is inversely proportional to the square root of gravity, which is lower on the moon.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$ .

Step 2: The acceleration due to gravity on the moon ( $g_m$ ) is approximately $\frac{1}{6}$ of the acceleration due to gravity on Earth ( $g_e$ ).

Step 3: Let $T_e$ be the time period on Earth and $T_m$ be the time period on the moon. Then, $T_e = 2\pi \sqrt{\frac{l}{g_e}}$ and $T_m = 2\pi \sqrt{\frac{l}{g_m}}$ .

Step 4: Since $g_m < g_e$ , it follows that $T_m > T_e$ .

Why other options are wrong:
- Option A: Less than $T$ . Incorrect, the time period will increase.
- Option B: Equal to $T$ . Incorrect, the time period will change due to the change in gravity.
- Option D: None of these. Incorrect, as option C is correct.

Key Concepts:
- Time period of a simple pendulum
- Acceleration due to gravity

Memory Aids:
- Lower gravity means a longer period.

Quick Tips:
- Remember the relationship between time period and gravity.

27 Easy

27 of 54Question permalink

Analyzing +4 | -1

Which of the following quantities is always negative in SHM?
Here,

\mathbf{s}

is displacement from mean position.

View answer and explanation

Correct option: C

\mathbf{a} \cdot \mathbf{s}

Option C (

\mathbf{a} \cdot \mathbf{s}

) is correct because acceleration and displacement are always in opposite directions in SHM.

Solution approach

Correct Answer + Core Reason: $\mathbf{a} \cdot \mathbf{s}$ . In SHM, acceleration and displacement are always in opposite directions.

Step-by-Step Solution:

Step 1: Recall the relationship between acceleration and displacement in SHM: $a = -\omega^2 s$ , where $a$ is the acceleration, $\omega$ is the angular frequency, and $s$ is the displacement from the mean position.

Step 2: The dot product of acceleration and displacement is given by $\mathbf{a} \cdot \mathbf{s} = |a||s|\cos\theta$ , where $\theta$ is the angle between the acceleration and displacement vectors.

Step 3: Since $a = -\omega^2 s$ , the acceleration and displacement are always in opposite directions. Therefore, the angle between them is $180^\circ$ or $\pi$ radians.

Step 4: Calculate the dot product: $\mathbf{a} \cdot \mathbf{s} = |a||s|\cos(180^\circ) = -|a||s|$ . Since $|a|$ and $|s|$ are always positive, the dot product is always negative.

Why other options are wrong:
- **Option A ( $\mathbf{F} \cdot \mathbf{a}$ ): The force and acceleration are in the same direction, so their dot product is always positive.
- Option B ( $\mathbf{v} \cdot \mathbf{s}$ ): The velocity and displacement can be in the same or opposite directions depending on the position and direction of motion, so their dot product can be positive or negative.
- Option D ( $\mathbf{F} \cdot \mathbf{v}$ ): The force and velocity can be in the same or opposite directions depending on the position and direction of motion, so their dot product can be positive or negative.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Acceleration in SHM
- Displacement in SHM
- Dot product of vectors

Memory Aids:
- Remember that in SHM, acceleration is always directed towards the mean position, opposite to the displacement.

Quick Tips:**
- When dealing with dot products in SHM, consider the directions of the vectors involved.

28 Easy

28 of 54Question permalink

Understanding +4 | -1

Two particles are executing simple harmonic motion. At an instant of time

t

, their displacements are

y_1=A \cos (\omega t)

and

y_2=A \sin (\omega t)

Then, the phase difference between

y_1

and

y_2

is

View answer and explanation

Correct option: B

90^{\circ}

Option B (

90^{\circ}

) is correct because the phase difference between

\cos(\omega t)

and

\sin(\omega t)

is

90^{\circ}

or

\pi/2

radians.

Solution approach

Correct Answer + Core Reason: $90^{\circ}$ . The phase difference between sine and cosine functions is $90^{\circ}$ .

Step-by-Step Solution:

Step 1: Given equations: $y_1 = A \cos(\omega t)$ and $y_2 = A \sin(\omega t)$ .

Step 2: Rewrite the sine function as a cosine function: $y_2 = A \cos(\omega t - \frac{{\pi}}{{2}})$ .

Step 3: Find the phase difference: The phase of $y_1$ is 0, and the phase of $y_2$ is $-\frac{{\pi}}{{2}}$ . The phase difference is $0 - (-\frac{{\pi}}{{2}}) = \frac{{\pi}}{{2}}$ .

Step 4: Convert radians to degrees: $\frac{{\pi}}{{2}} \text{ radians} = 90^{\circ}$ .

Step 5: Conclusion: The phase difference between $y_1$ and $y_2$ is $90^{\circ}$ .

Why other options are wrong:
- Option A: Incorrect. The phase difference is not $120^{\circ}$ .
- Option C: Incorrect. The phase difference is not $180^{\circ}$ .
- Option D: Incorrect. The phase difference is not zero.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase Difference
- Sine and Cosine Functions

Memory Aids:
- Remember that $\sin(\theta) = \cos(\theta - 90^{\circ})$ .

Quick Tips:
- Convert sine functions to cosine functions (or vice versa) to easily find the phase difference.

29 Easy

29 of 54Question permalink

Application +4 | -1

A particle is attached to a vertical spring and is pulled down a distance

0.04 \mathrm{~m}

below its equilibrium position and is released from rest. The initial upward acceleration of the particle is

0.30 \mathrm{~ms}^{-2}

. The period of the oscillation is

View answer and explanation

Correct option: D

2.29 \mathrm{~s}

Option D (

2.29 \mathrm{~s}

) is correct because it is calculated using the given acceleration and displacement to find the angular frequency and then the period.

Solution approach

Correct Answer + Core Reason: $2.29 \mathrm{~s}$ . The period is calculated from the initial acceleration and displacement.

Step-by-Step Solution:

Step 1: Identify the given information: displacement $x = 0.04 \mathrm{~m}$ , acceleration $a = 0.30 \mathrm{~ms}^{-2}$ .

Step 2: Recall the relationship between acceleration and displacement in SHM: $a = \omega^2 x$ , where $\omega$ is the angular frequency.

Step 3: Calculate the angular frequency: $\omega = \sqrt{\frac{a}{x}} = \sqrt{\frac{0.30}{0.04}} = \sqrt{7.5} \approx 2.738 \mathrm{~rad/s}$

Step 4: Calculate the period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2.738} \approx 2.29 \mathrm{~s}$

Why other options are wrong:
- Option A: Incorrect. This value doesn't match the correct calculation using the given acceleration and displacement.
- Option B: Incorrect. This value doesn't match the correct calculation using the given acceleration and displacement.
- Option C: Incorrect. This value doesn't match the correct calculation using the given acceleration and displacement.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Relationship between acceleration, displacement, and angular frequency in SHM: $a = \omega^2 x$
- Period of SHM: $T = \frac{2\pi}{\omega}$

Memory Aids:
- Remember the formula $a = \omega^2 x$ and $T = \frac{2\pi}{\omega}$ .

Quick Tips:
- Always check the units of the given values before plugging them into the formulas.
- Make sure to use the correct relationship between acceleration, displacement, and angular frequency.

30 Easy

30 of 54Question permalink

Application +4 | -1

The time period of oscillation of a simple pendulum of length

L

suspended from the roof a vehicle, which moves without friction down an inclined plane of inclination

\alpha

, is given by:

View answer and explanation

Correct option: A

2 \pi \sqrt{L /(g \cos \alpha)}

Option A (

2 \pi \sqrt{L /(g \cos \alpha)}

) is correct because the effective gravitational acceleration is

g \cos \alpha

when the vehicle moves down the inclined plane.

Solution approach

Correct Answer + Core Reason: $2 \pi \sqrt{L /(g \cos \alpha)}$ . The effective gravitational acceleration is $g \cos \alpha$ due to the inclined plane.

Step-by-Step Solution:

Step 1: Identify the effective gravitational acceleration. When the vehicle moves down the inclined plane, the component of gravity acting along the direction perpendicular to the string is $g \cos \alpha$ .

Step 2: Recall the formula for the time period of a simple pendulum: $T = 2 \pi \sqrt{\frac{L}{g_{eff}}}$ , where $L$ is the length of the pendulum and $g_{eff}$ is the effective gravitational acceleration.

Step 3: Substitute $g_{eff} = g \cos \alpha$ into the time period formula: $T = 2 \pi \sqrt{\frac{L}{g \cos \alpha}}$ .

Step 4: Final Answer: $T = 2 \pi \sqrt{\frac{L}{g \cos \alpha}}$

Why other options are wrong:
- Option B: This option uses $g \sin \alpha$ , which is the component of gravity along the inclined plane, not perpendicular to the string.
- Option C: This option uses $g$ , which is the gravitational acceleration in the absence of any inclination. This is incorrect because the inclined plane affects the effective gravitational acceleration.
- Option D: This option uses $g \tan \alpha$ , which is not relevant to the effective gravitational acceleration in this scenario.

Key Concepts:
- Simple pendulum
- Effective gravitational acceleration
- Inclined plane

Memory Aids:
- Remember that the effective gravity is the component perpendicular to the string.

Quick Tips:
- Always consider the effective gravitational acceleration when dealing with pendulums on inclined planes or in accelerating frames.

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