OSCILLATIONS MCQs for NEET UG PHYSICS — Hard

1 Hard

Application +4 | -1

The phase difference (in radian) between two SHM whose equations are

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

and

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

at

t=0.5 \mathrm{~s}

is:

View answer and explanation

Correct option: B

\frac{13 \pi}{12}

Option B (

\frac{{13 \pi}}{{12}}

) is correct because it represents the phase difference calculated at t=0.5s between the two given SHM equations.

Solution approach

Correct Answer + Core Reason: $\frac{{13 \pi}}{{12}}$ . The phase difference is calculated by substituting t=0.5s into the arguments of the sine functions and finding the difference.

Step-by-Step Solution:

Step 1: Identify the given equations: $y_1 = 10 \sin(10 \pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8 \pi t + \frac{\pi}{4})$ .

Step 2: Calculate the phase of $y_1$ at $t = 0.5 \text{{s}}$ : $\phi_1 = 10 \pi (0.5) + \frac{\pi}{3} = 5\pi + \frac{\pi}{3} = \frac{{16\pi}}{3}$ .

Step 3: Calculate the phase of $y_2$ at $t = 0.5 \text{{s}}$ : $\phi_2 = 8 \pi (0.5) + \frac{\pi}{4} = 4\pi + \frac{\pi}{4} = \frac{{17\pi}}{4}$ .

Step 4: Find the phase difference: $\Delta \phi = |\phi_1 - \phi_2| = |\frac{{16\pi}}{3} - \frac{{17\pi}}{4}| = |\frac{{64\pi - 51\pi}}{12}| = |\frac{{13\pi}}{12}| = \frac{{13\pi}}{12}$ .

Why other options are wrong:
- **Option A ( $\frac{{11 \pi}}{{12}}$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s.
- Option C ( $\pi$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s.
- Option D ( $\frac{{17 \pi}}{{12}}$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s. The absolute value of the difference must be taken.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase difference calculation

Memory Aids:
- Phase is the argument of the sine or cosine function.

Quick Tips:**
- Always substitute the given time into the phase equations before finding the difference.
- Remember to take the absolute value of the phase difference.

2 Hard

2 of 14Question permalink

Application +4 | -1

Maximum kinetic energy of a particle of mass

1 \mathrm{~kg}

in SHM is

8 \mathrm{~J}

. Time period of SHM is

4 \mathrm{~s}

. Maximum potential energy during the motion is

10 \mathrm{~J}

. Then,

View answer and explanation

Correct option: C

maximum acceleration of the particle is approximately

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

Option C (maximum acceleration of the particle is approximately

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

) is correct because it is calculated using the angular frequency and maximum kinetic energy to find the amplitude and then the maximum acceleration.

Solution approach

Correct Answer + Core Reason: maximum acceleration of the particle is approximately $6.3 \mathrm{~ms}^{-2}$ . The maximum acceleration is calculated from the amplitude and angular frequency.

Step-by-Step Solution:

Step 1: Given: mass $m = 1 \mathrm{~kg}$ , maximum kinetic energy $K_{max} = 8 \mathrm{~J}$ , time period $T = 4 \mathrm{~s}$ .

Step 2: The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \mathrm{~rad/s}$ .

Step 3: The maximum kinetic energy is given by $K_{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (A\omega)^2$ , where $A$ is the amplitude.

Step 4: Solve for the amplitude: $A = \sqrt{\frac{2K_{max}}{m\omega^2}} = \sqrt{\frac{2(8)}{1(\frac{\pi}{2})^2}} = \sqrt{\frac{16}{\frac{\pi^2}{4}}} = \sqrt{\frac{64}{\pi^2}} \approx \sqrt{\frac{64}{9.87}} \approx \sqrt{6.48} \approx 2.55 \mathrm{~m}$ .

Step 5: The maximum acceleration is given by $a_{max} = A\omega^2 = 2.55 \times (\frac{\pi}{2})^2 = 2.55 \times \frac{\pi^2}{4} \approx 2.55 \times \frac{9.87}{4} \approx 2.55 \times 2.47 \approx 6.3 \mathrm{~ms}^{-2}$ .

Why other options are wrong:
- Option A: Incorrect. The amplitude is approximately 2.55 m, not 4.53 m.
- Option B: Incorrect. The minimum potential energy is not necessarily 4 J. The problem does not provide enough information to determine the minimum potential energy. We only know the *maximum* potential energy is 10J.
- Option D: Incorrect. The minimum kinetic energy is 0 J, not 2 J.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency: $\omega = \frac{2\pi}{T}$
- Maximum kinetic energy: $K_{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (A\omega)^2$
- Maximum acceleration: $a_{max} = A\omega^2$

Memory Aids:
- Remember the formulas for angular frequency, maximum kinetic energy, and maximum acceleration in SHM.

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

3 Hard

3 of 14Question permalink

Understanding +4 | -1

The equation of a damped simple harmonic motion is

m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0

. Then, the angular frequency of oscillation is

View answer and explanation

Correct option: A

\omega=\left(\frac{k}{m}-\frac{b^2}{4 m^2}\right)^{1 / 2}

Option A (

\omega=\left(\frac{k}{m}-\frac{b^2}{4 m^2}\right)^{1 / 2}

) is correct because it represents the angular frequency of a damped harmonic oscillator, derived from solving the given differential equation.

Solution approach

Correct Answer + Core Reason: $\omega=\left(\frac{k}{m}-\frac{b^2}{4 m^2}\right)^{1 / 2}$ . This is the standard formula for the angular frequency of a damped harmonic oscillator.

Step-by-Step Solution:

Step 1: Identify the given equation: $m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0$

Step 2: Recall the general solution for a damped harmonic oscillator: $x(t) = A e^{-\gamma t} \cos(\omega t + \phi)$ , where $\gamma = \frac{b}{2m}$ is the damping coefficient and $\omega$ is the angular frequency.

Step 3: The angular frequency $\omega$ is given by: $\omega = \sqrt{\omega_0^2 - \gamma^2}$ , where $\omega_0 = \sqrt{\frac{k}{m}}$ is the natural angular frequency.

Step 4: Substitute the values of $\omega_0$ and $\gamma$ into the equation: $\omega = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2} = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$

Step 5: Therefore, the angular frequency of oscillation is $\omega=\left(\frac{k}{m}-\frac{b^2}{4 m^2}\right)^{1 / 2}$ .

Why other options are wrong:
- Option B: $\omega=\left(\frac{k}{m}-\frac{b}{4 m}\right)^{1 / 2}$ is incorrect because it has 'b' instead of 'b²' in the expression and lacks the square in the denominator.
- Option C: $\omega=\left(\frac{k}{m}-\frac{b^2}{4 m}\right)^{1 / 2}$ is incorrect because it has '4m' instead of '4m²' in the denominator.
- Option D: $\omega=\left(\frac{k}{m}-\frac{b^2}{4 m^2}\right)^{-1 / 2}$ is incorrect because it is the inverse of the correct angular frequency.

Key Concepts:
- Damped Harmonic Motion
- Angular Frequency
- Damping Coefficient

Memory Aids:
- Remember the formula for angular frequency in damped harmonic motion: $\omega = \sqrt{\omega_0^2 - \gamma^2}$

Quick Tips:
- Identify the coefficients in the given differential equation correctly.
- Use the correct formula for angular frequency in damped harmonic motion.

4 Hard

4 of 14Question permalink

Analyzing +4 | -1

Identify the correct statement.

View answer and explanation

Correct option: D

All statements are correct

Option D (All statements are correct) is correct because all the given statements about the time period of a simple pendulum, a ball in a tunnel, and a satellite are accurate.

Solution approach

Correct Answer + Core Reason: All statements are correct. Each statement accurately describes the time period of the respective system.

Step-by-Step Solution:

Step 1: Analyze statement A: Simple pendulum in a lift moving upward with constant speed. The time period is $T=2 \pi \sqrt{\frac{L}{g}}$ . Since the lift moves with constant speed, there is no pseudo force, and the effective acceleration is still $g$ . Thus, the time period remains the same.

Step 2: Analyze statement B: Ball released into a tunnel dug through the diameter of the earth. The time period is $T=2 \pi \sqrt{\frac{R}{g}}$ . The acceleration of the ball is $a = \frac{g}{R}x$ , where $x$ is the displacement from the center. Thus, $\omega = \sqrt{\frac{g}{R}}$ , and $T = \frac{2\pi}{\omega} = 2 \pi \sqrt{\frac{R}{g}}$ .

Step 3: Analyze statement C: Satellite revolving very close to the surface of the earth. The time period is $T=2 \pi \sqrt{\frac{R}{g}}$ . The orbital velocity is $v = \sqrt{\frac{GM}{R}} = \sqrt{gR}$ . The time period is $T = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{gR}} = 2 \pi \sqrt{\frac{R}{g}}$ .

Step 4: Since all statements are correct, the answer is D.

Why other options are wrong:
- Option A: If simple pendulum of small length $L$ is suspended from the ceiling of a lift which is moving upward with constant speed, then its time period $T=2 \pi \sqrt{\frac{L}{g}}$ : This statement is correct, but not the only correct statement.
- Option B: If the tunnel is dug through the diameter of the earth, and ball is released into it from earth surface, ball will do SHM of time period $T=2 \pi \sqrt{\frac{R}{g}},$ where $R$ is radius of earth: This statement is correct, but not the only correct statement.
- Option C: Satellite revolving very close to the surface of the earth has time period $T=2 \pi \sqrt{\frac{R}{g}},$ where $R$ is radius of earth: This statement is correct, but not the only correct statement.

Key Concepts:
- Simple pendulum
- SHM in a tunnel through the Earth
- Satellite motion

Memory Aids:
- Time period formulas for different systems.

Quick Tips:
- Remember the conditions for SHM and the relevant formulas.

5 Hard

5 of 14Question permalink

Analyzing +4 | -1

I. Average kinetic energy in one oscillation during SHM of a body is

\frac{1}{4} m \omega^2 A^2

.

II. Maximum kinetic energy is

\frac{1}{2} m \omega^2 A^2

.

Which of the following statement(s) is/are correct?

View answer and explanation

Correct option: C

Both I and II

Option C (Both I and II) is correct because both the average kinetic energy and the maximum kinetic energy during SHM are correctly stated.

Solution approach

Correct Answer + Core Reason: Both I and II. Both statements about average and maximum kinetic energy in SHM are correct.

Step-by-Step Solution:

Step 1: Statement I: Average kinetic energy in one oscillation during SHM of a body is $\frac{1}{4} m \omega^2 A^2$ . This is true. The average kinetic energy is half of the maximum kinetic energy, and the maximum kinetic energy is $\frac{1}{2} m \omega^2 A^2$ . Therefore, the average kinetic energy is $\frac{1}{2} \times \frac{1}{2} m \omega^2 A^2 = \frac{1}{4} m \omega^2 A^2$ .

Step 2: Statement II: Maximum kinetic energy is $\frac{1}{2} m \omega^2 A^2$ . This is true. The maximum kinetic energy occurs when the potential energy is zero, which is at the equilibrium position. At this point, the total energy is equal to the kinetic energy, and the total energy is $\frac{1}{2} m \omega^2 A^2$ .

Why other options are wrong:
- Option A: Only I. This is incorrect because Statement II is also correct.
- Option B: Only II. This is incorrect because Statement I is also correct.
- Option D: Neither I nor II. This is incorrect because both statements are correct.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Kinetic Energy in SHM
- Potential Energy in SHM
- Average Kinetic Energy
- Maximum Kinetic Energy

Memory Aids:
- Remember that the total energy in SHM is $E = \frac{1}{2} m \omega^2 A^2$ .
- Remember that the average kinetic energy is half of the maximum kinetic energy.

Quick Tips:
- Understand the energy variations in SHM.
- Know the formulas for average and maximum kinetic energy.

6 Hard

6 of 14Question permalink

Application +4 | -1

The velocities of a particle in SHM at positions

x_1

and

x_2

are

v_1

and

v_2,

respectively, its time period will be

View answer and explanation

Correct option: C

2 \pi \sqrt{\frac{\left(x_1^2-x_2^2\right)}{\left(v_2^2-v_1^2\right)}}

Option C (

2 \pi \sqrt{\frac{\left(x_1^2-x_2^2\right)}{\left(v_2^2-v_1^2\right)}}

) is correct because the time period is derived from the relationship between velocity, position, and angular frequency in SHM.

Solution approach

Correct Answer + Core Reason: $2 \pi \sqrt{\frac{\left(x_1^2-x_2^2\right)}{\left(v_2^2-v_1^2\right)}}$ . The time period is derived from the velocities and positions at two different points in SHM.

Step-by-Step Solution:

Step 1: Recall the relationship between velocity and displacement in SHM: $v = \omega \sqrt{A^2 - x^2}$ , where $v$ is the velocity, $\omega$ is the angular frequency, $A$ is the amplitude, and $x$ is the displacement.

Step 2: Write the equations for the two given positions and velocities:
$v_1 = \omega \sqrt{A^2 - x_1^2}$ and $v_2 = \omega \sqrt{A^2 - x_2^2}$ .

Step 3: Square both equations:
$v_1^2 = \omega^2 (A^2 - x_1^2)$ and $v_2^2 = \omega^2 (A^2 - x_2^2)$ .

Step 4: Divide both equations by $\omega^2$ :
$\frac{v_1^2}{\omega^2} = A^2 - x_1^2$ and $\frac{v_2^2}{\omega^2} = A^2 - x_2^2$ .

Step 5: Subtract the second equation from the first:
$\frac{v_1^2}{\omega^2} - \frac{v_2^2}{\omega^2} = (A^2 - x_1^2) - (A^2 - x_2^2) = x_2^2 - x_1^2$ .

Step 6: Rearrange to solve for $\omega^2$ :
$\omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}$ .

Step 7: Solve for $\omega$ :
$\omega = \sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}}$ .

Step 8: Recall the relationship between angular frequency and time period: $T = \frac{2\pi}{\omega}$ .

Step 9: Substitute the expression for $\omega$ :\nT = $\frac{2\pi}{\sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}}} = 2 \pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} = 2 \pi \sqrt{\frac{x_1^2 - x_2^2}{v_2^2 - v_1^2}}$ .

Why other options are wrong:
- **Option A ( $2 \pi \sqrt{\frac{\left(v_1^2-v_2^2\right)}{\left(x_2^2-x_1^2\right)}}$ ): This is incorrect because it has the inverse relationship between velocities and positions.
- Option B ( $2 \pi \sqrt{\frac{\left(x_1^2+x_2^2\right)}{\left(v_2^2-v_1^2\right)}}$ ): This is incorrect because it uses the sum of squares of positions instead of the difference.
- Option D ( $2 \pi \sqrt{\frac{\left(x_1^2+x_2^2\right)}{\left(v_2^2+v_1^2\right)}}$ ): This is incorrect because it uses the sum of squares for both positions and velocities.

Key Concepts:**
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Angular frequency ( $\omega$ )
- Time period ( $T$ )

Memory Aids:
- Remember the relationship $v = \omega \sqrt{A^2 - x^2}$ and $T = \frac{2\pi}{\omega}$ .

Quick Tips:
- When dealing with SHM problems involving velocities and positions, use the relationship $v = \omega \sqrt{A^2 - x^2}$ .

7 Hard

7 of 14Question permalink

Application +4 | -1

A solid cube floats in water half immersed and has small vertical oscillations of time period

\frac{\pi}{5} \mathrm{~s}

. Its mass (in kg) is (Take,

g=10 \mathrm{~ms}^{-2}

)

View answer and explanation

Correct option: A

4

Option A (4) is correct because the mass is calculated from the time period of small vertical oscillations of a floating cube.

Solution approach

Correct Answer + Core Reason: 4. The mass is calculated from the time period of small vertical oscillations.

Step-by-Step Solution:

Step 1: The cube is half immersed, so the weight of the cube equals the weight of the water displaced. Let the side of the cube be $a$ . Then, $mg = a^2 \times \frac{a}{2} \times \rho_w \times g$ , where $\rho_w$ is the density of water (1000 kg/m³).

Step 2: When the cube is displaced by a small distance $x$ downwards, an extra buoyant force acts upwards. This force is $F = a^2 x \rho_w g$ .

Step 3: This force provides the restoring force for SHM. So, $F = -kx$ , where $k = a^2 \rho_w g$ .

Step 4: The time period of oscillation is $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{a^2 \rho_w g}}$ .

Step 5: Given $T = \frac{\pi}{5}$ , so $\frac{\pi}{5} = 2\pi \sqrt{\frac{m}{a^2 \rho_w g}}$ .

Step 6: Squaring both sides, we get $\frac{\pi^2}{25} = 4\pi^2 \frac{m}{a^2 \rho_w g}$ .

Step 7: Simplifying, we get $\frac{1}{100} = \frac{m}{a^2 \rho_w g}$ , so $m = \frac{a^2 \rho_w g}{100}$ .

Step 8: From Step 1, $mg = \frac{a^3 \rho_w g}{2}$ , so $m = \frac{a^3 \rho_w}{2}$ .

Step 9: Substituting $m$ from Step 7, we get $\frac{a^2 \rho_w g}{100} = \frac{a^3 \rho_w}{2}$ .

Step 10: Simplifying, we get $\frac{g}{100} = \frac{a}{2}$ , so $a = \frac{2g}{100} = \frac{20}{100} = 0.2 \text{ m}$ .

Step 11: Substituting $a$ into $m = \frac{a^3 \rho_w}{2}$ , we get $m = \frac{(0.2)^3 \times 1000}{2} = \frac{0.008 \times 1000}{2} = \frac{8}{2} = 4 \text{ kg}$ .

Why other options are wrong:
- Option B: 0 This is incorrect. The mass cannot be zero.
- Option C: 0 This is incorrect. The mass cannot be zero.
- Option D: 0 This is incorrect. The mass cannot be zero.

Key Concepts:
- Buoyancy
- Simple Harmonic Motion (SHM)
- Time Period of SHM

Memory Aids:
- Remember the formula for the time period of SHM: $T = 2\pi \sqrt{\frac{m}{k}}$ .

Quick Tips:
- Relate the restoring force to the displacement to find the spring constant $k$ .
- Use the equilibrium condition to relate the mass and the dimensions of the cube.

8 Hard

8 of 14Question permalink

Application +4 | -1

A uniform rod of length

l

is suspended from any point

P

. It is then made to undergo small oscillations. If

P

is the

View answer and explanation

Correct option: B

end point,

T

is

2 \pi \sqrt{\frac{l}{3 g}}

Option B (end point,

T

is

2 \pi \sqrt{\frac{{l}}{{3 g}}}

) is correct because the time period of a physical pendulum depends on the moment of inertia and the distance from the pivot point to the center of mass.

Solution approach

Correct Answer + Core Reason: end point, $T$ is $2 \pi \sqrt{\frac{l}{3 g}}$ . The time period of a physical pendulum is given by $T = 2 \pi \sqrt{\frac{I}{m g d}}$ , where $I$ is the moment of inertia about the pivot point, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance from the pivot to the center of mass.

Step-by-Step Solution:

Step 1: Identify the given information: A uniform rod of length $l$ is suspended from a point $P$ .

Step 2: Determine the moment of inertia $I$ about the end point. For a uniform rod, $I = \frac{1}{3} m l^2$ .

Step 3: Determine the distance $d$ from the pivot point to the center of mass. In this case, $d = \frac{l}{2}$ .

Step 4: Substitute the values of $I$ and $d$ into the formula for the time period: $T = 2 \pi \sqrt{\frac{\frac{1}{3} m l^2}{m g (\frac{l}{2})}}$ .

Step 5: Simplify the expression: $T = 2 \pi \sqrt{\frac{\frac{1}{3} l^2}{g \frac{l}{2}}} = 2 \pi \sqrt{\frac{2l}{3g}}$ .

Final Answer: The time period of oscillation is $T = 2 \pi \sqrt{\frac{2l}{3g}}$ . Note: There was a typo in the original answer. The correct answer should be $2 \pi \sqrt{\frac{2l}{3g}}$ .

Why other options are wrong:
- **Option A (centre of mass, $T$ is zero): This is incorrect because if the rod is suspended from its center of mass, it will not oscillate. The time period is undefined, not zero.
- Option C (centre of mass, $T=2 \pi \sqrt{\frac{l}{g}}$ ): This is incorrect because if the rod is suspended from its center of mass, it will not oscillate. The time period is undefined.
- Option D (end point, $T$ is infinite): This is incorrect because the time period is finite and can be calculated using the formula for a physical pendulum.

Key Concepts:
- Physical Pendulum
- Moment of Inertia
- Time Period of Oscillation

Memory Aids:**
- Remember the formula for the time period of a physical pendulum: $T = 2 \pi \sqrt{\frac{I}{m g d}}$ .

Quick Tips:
- Always determine the moment of inertia about the pivot point.
- Identify the distance from the pivot point to the center of mass.

9 Hard

9 of 14Question permalink

Application +4 | -1

A block whose mass is

2 \mathrm{~kg}

is fastened to a spring whose spring constant is

100 \mathrm{Nm}^{-1}

. It is pulled to a distance of

0.1 \mathrm{~m}

from over a frictionless surface and is released at

t=0

. Calculate the kinetic energy of the block when it is

0.05 \mathrm{~m}

away from its mean position.

View answer and explanation

Correct option: B

0.37 \mathrm{~J}

Option B (

0.37 \mathrm{~J}

) is correct because it is the kinetic energy calculated using the conservation of energy principle for SHM.

Solution approach

Correct Answer + Core Reason: $0.37 \mathrm{~J}$ . The kinetic energy is calculated using the conservation of energy principle.

Step-by-Step Solution:

Step 1: Identify the given information: mass $m = 2 \mathrm{~kg}$ , spring constant $k = 100 \mathrm{Nm}^{-1}$ , amplitude $A = 0.1 \mathrm{~m}$ , and displacement $x = 0.05 \mathrm{~m}$ .

Step 2: Recall the formula for total energy in SHM: $E = \frac{1}{2} k A^2$ .

Step 3: Recall the formula for potential energy at displacement x: $U = \frac{1}{2} k x^2$ .

Step 4: Recall the formula for kinetic energy: $K = E - U$ .

Step 5: Calculate the total energy: $E = \frac{1}{2} (100 \mathrm{Nm}^{-1}) (0.1 \mathrm{~m})^2 = 0.5 \mathrm{~J}$ .

Step 6: Calculate the potential energy: $U = \frac{1}{2} (100 \mathrm{Nm}^{-1}) (0.05 \mathrm{~m})^2 = 0.125 \mathrm{~J}$ .

Step 7: Calculate the kinetic energy: $K = E - U = 0.5 \mathrm{~J} - 0.125 \mathrm{~J} = 0.375 \mathrm{~J} \approx 0.37 \mathrm{~J}$ .

Why other options are wrong:
- **Option A ( $3 \mathrm{~J}$ ): This value is incorrect because it likely involves a miscalculation or misunderstanding of the energy formulas in SHM.
- Option C ( $4.16 \mathrm{~J}$ ): This value is incorrect and doesn't correspond to any logical calculation using the given parameters and SHM principles.
- Option D (Zero):** This is incorrect because the block has a non-zero velocity at $x = 0.05 \mathrm{~m}$ , hence non-zero kinetic energy.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Conservation of Energy
- Potential Energy in SHM: $U = \frac{1}{2} k x^2$
- Total Energy in SHM: $E = \frac{1}{2} k A^2$

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

Quick Tips:
- Always identify the given parameters and the required quantity.
- Use the correct formulas for potential and total energy in SHM.
- Apply the conservation of energy principle to find the kinetic energy.

10 Hard

10 of 14Question permalink

Application +4 | -1

A uniform spring whose unstretched length is

l

has a force constant

k

. The spring is cut into two pieces of unstretched lengths

l_1

and

l_2

, where

l_1=n l_2

and

n

is being an integer. Now, a mass

m

is made to oscillate with first spring. The time period of its oscillation would be

View answer and explanation

Correct option: A

T=2 \pi \sqrt{\frac{m n}{k(n+1)}}

Option A,

T=2 \pi \sqrt{\frac{{m n}}{{k(n+1)}}}

is correct because the effective spring constant for the first spring is

k(n+1)/n

, leading to the given time period.

Solution approach

Correct Answer + Core Reason: $T=2 \pi \sqrt{\frac{{m n}}{{k(n+1)}}}$ The time period depends on the effective spring constant of the first spring after cutting.

Step-by-Step Solution:

Step 1: Determine the spring constant for each piece. Since $l_1 = n l_2$ , we have $l = l_1 + l_2 = n l_2 + l_2 = (n+1) l_2$ . Thus, $l_2 = \frac{l}{n+1}$ and $l_1 = \frac{n l}{n+1}$ .

Step 2: The spring constant is inversely proportional to the length of the spring. Therefore, ${k_1}{l_1} = k l$ and ${k_2}{l_2} = k l$ .

Step 3: Find the spring constant of the first spring ( $k_1$ ): ${k_1} = \frac{{k l}}{{{l_1}}} = \frac{{k l}}{{\frac{{n l}}{{n+1}}}} = \frac{{k(n+1)}}{n}$ .

Step 4: Write the expression for the time period of oscillation: $T = 2 \pi \sqrt{\frac{m}{k}}$ .

Step 5: Substitute the spring constant of the first spring ( $k_1$ ) into the time period equation: $T = 2 \pi \sqrt{\frac{m}{{{k_1}}}} = 2 \pi \sqrt{\frac{m}{{\frac{{k(n+1)}}{n}}}} = 2 \pi \sqrt{\frac{{m n}}{{k(n+1)}}}$ .

Why other options are wrong:
- Option B: $T=2 \pi \sqrt{\frac{m}{{k(n+1)}}}$ This is incorrect because it doesn't account for the correct spring constant of the first spring after cutting. It misses the factor of 'n' in the numerator.
- Option C: $T=2 \pi \sqrt{\frac{m}{{n k}}}$ This is incorrect because it uses the inverse of the correct spring constant. It also doesn't account for the (n+1) factor.
- Option D: $T=2 \pi \sqrt{\frac{{m(n+1)}}{{n k}}}$ This is incorrect because it uses the inverse of the correct spring constant. It has the (n+1) factor in the numerator instead of the denominator.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Spring Constant
- Time Period of Oscillation

Memory Aids:
- $T = 2 \pi \sqrt{\frac{m}{k}}$ (Time period of a spring-mass system)
- Spring constant is inversely proportional to length.

Quick Tips:
- When a spring is cut into pieces, the spring constant of each piece changes.
- The spring constant is inversely proportional to the length of the spring.

11 Hard

11 of 14Question permalink

Analyzing +4 | -1

The equation of a travelling wave is

y=15 \sin ^2(3 x-4 t),

where

x, y

are in meter and

t

is in second. The amplitude of oscillation of particle is:

View answer and explanation

Correct option: B

7.5 m

Option B (7.5 m) is correct because the amplitude of oscillation is half of the coefficient due to the sin² term.

Solution approach

Correct Answer + Core Reason: 7.5 m. The amplitude is half the coefficient due to the $\sin^2$ term.

Step-by-Step Solution:

Step 1: Rewrite the equation using the trigonometric identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ : $y = 15 \sin^2(3x - 4t) = 15 \left( \frac{1 - \cos(2(3x - 4t))}{2} \right) = \frac{15}{2} - \frac{15}{2} \cos(6x - 8t)$ .

Step 2: Identify the amplitude. The amplitude of the oscillation is the coefficient of the cosine term, which is $\frac{15}{2} = 7.5 \text{{ m}}$ .

Why other options are wrong:
- Option A (15 m): This is incorrect because it doesn't account for the trigonometric identity.
- Option C (30 m): This is incorrect because it's twice the correct amplitude.
- Option D (10.3 m): This is incorrect and doesn't have a clear derivation from the given equation.

Key Concepts:
- Wave equation
- Trigonometric Identities
- Amplitude of a wave

Memory Aids:
- Remember the trigonometric identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ .

Quick Tips:
- Use trigonometric identities to simplify the wave equation before identifying the amplitude.

12 Hard

12 of 14Question permalink

Application +4 | -1

Two simple harmonic motion are represented by

{y_1} = 5\left( {\sin 2\pi t + \sqrt 3 \cos 2\pi t} \right)

{y_2} = 5\sin \left( {2\pi t + \frac{\pi }{4}} \right)

The ratio of the amplitudes of two SHM’s is

View answer and explanation

Correct option: C

2\mathrm{ }:1

Option C (2 : 1) is correct because the ratio of amplitudes is found by simplifying the first equation and comparing the amplitudes.

Solution approach

Correct Answer + Core Reason: $2\mathrm{ }:1$ . The ratio is found by calculating the amplitude of the first SHM and comparing it to the second.

Step-by-Step Solution:

Step 1: Identify the given equations: ${y_1} = 5\left( {\sin 2\pi t + \sqrt 3 \cos 2\pi t} \right)$ and ${y_2} = 5\sin \left( {2\pi t + \frac{\pi }{4}} \right)$ .

Step 2: Find the amplitude of the first SHM. Let $A_1 = 5\sqrt{1^2 + (\sqrt{3})^2} = 5\sqrt{1 + 3} = 5\sqrt{4} = 5 \times 2 = 10$ .

Step 3: Find the amplitude of the second SHM. $A_2 = 5$ .

Step 4: Calculate the ratio of the amplitudes: $\frac{A_1}{A_2} = \frac{10}{5} = 2$ .

Step 5: Express the ratio as $2:1$ .

Why other options are wrong:
- **Option A ( $1\mathrm{ }:1$ ): This is incorrect because it doesn't correctly calculate the amplitude of the first SHM.
- Option B ( $1\mathrm{ }:2$ ): This is incorrect because it inverts the correct ratio.
- Option D ( $1:\sqrt 3$ ): This is incorrect because it only considers the coefficient of the cosine term in the first equation.

Key Concepts:
- Superposition of SHMs
- Resultant amplitude
- Ratio of amplitudes

Memory Aids:**
- Remember the formula $A = \sqrt{A_1^2 + A_2^2}$ for finding the resultant amplitude of two SHMs with the same frequency.

Quick Tips:
- When comparing SHMs, always calculate the amplitudes correctly before finding the ratio.

13 Hard

13 of 14Question permalink

Application +4 | -1

In a damped oscillation with damping constant

b,

the time taken for its mechanical energy to drop
to half its value is

View answer and explanation

Correct option: A

\frac{m}{b}\ln 2

Option A (

\frac{m}{b}\ln 2

) is correct because it represents the time taken for the mechanical energy of a damped oscillator to halve, derived from the energy decay equation.

Solution approach

Correct Answer + Core Reason: $\frac{m}{b}\ln 2$ . This is the time taken for the mechanical energy to drop to half its value in a damped oscillation.

Step-by-Step Solution:

Step 1: The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-\frac{bt}{m}}$ , where $E_0$ is the initial energy, $b$ is the damping constant, $m$ is the mass, and $t$ is time.

Step 2: We want to find the time $t$ when $E(t) = \frac{1}{2} E_0$ .

Step 3: Substitute this into the energy equation: $\frac{1}{2} E_0 = E_0 e^{-\frac{bt}{m}}$ .

Step 4: Divide both sides by $E_0$ : $\frac{1}{2} = e^{-\frac{bt}{m}}$ .

Step 5: Take the natural logarithm of both sides: $\ln(\frac{1}{2}) = -\frac{bt}{m}$ .

Step 6: Since $\ln(\frac{1}{2}) = -\ln(2)$ , we have $-\ln(2) = -\frac{bt}{m}$ .

Step 7: Solve for $t$ : $t = \frac{m}{b} \ln 2$ .

Why other options are wrong:
- **Option B ( $\frac{2m}{b}\ln 2$ ): This is incorrect because the factor of 2 is not present in the correct derivation of the time.
- Option C ( $\frac{b}{m}\ln 2$ ):** This is incorrect because the ratio of $m$ and $b$ is inverted.
- **Option D ( $\frac{b}{2m}\ln 2$ ):** This is incorrect because the ratio of $m$ and $b$ is inverted, and there is an extra factor of 2 in the denominator.

Key Concepts:
- Damped Oscillations
- Exponential Decay of Energy
- Natural Logarithm

Memory Aids:
- Remember the energy decay equation $E(t) = E_0 e^{-\frac{bt}{m}}$ .

Quick Tips:
- Always start with the correct equation for energy decay in damped oscillations.
- Use natural logarithms to solve for time in exponential equations.

14 Hard

14 of 14Question permalink

Understanding +4 | -1

The displacement, velocity and acceleration in a simple harmonic motion are related as the

View answer and explanation

Correct option: A

displacement, velocity and acceleration all act in the same direction

Option D (displacement and acceleration are anti-parallel and both perpendicular to the velocity) is incorrect. Displacement and acceleration are anti-parallel, but they are not necessarily perpendicular to velocity. The correct option is that displacement and acceleration are anti-parallel and both perpendicular to the velocity.

Solution approach

Correct Answer + Core Reason: displacement and acceleration are anti-parallel and both perpendicular to the velocity. In SHM, acceleration is always directed towards the mean position, opposite to the displacement.

Step-by-Step Solution:

Step 1: Recall the relationship between displacement ( $x$ ) and acceleration ( $a$ ) in SHM: $a = -\omega^2 x$ , where $\omega$ is the angular frequency.

Step 2: The negative sign indicates that the acceleration is always in the opposite direction to the displacement. Therefore, displacement and acceleration are anti-parallel.

Step 3: Velocity is maximum at the mean position (zero displacement) and zero at the extreme positions (maximum displacement). Velocity is perpendicular to both displacement and acceleration at the extreme positions and parallel at the mean position. However, the question is asking for a general relationship, and at the extreme positions, displacement and acceleration are anti-parallel and velocity is momentarily zero.

Why other options are wrong:
- Option A: displacement, velocity and acceleration all act in the same direction. Incorrect, displacement and acceleration are in opposite directions.
- Option B: displacement and velocity act in the same direction but acceleration in the opposite direction. Incorrect, displacement and acceleration are always in opposite directions.
- Option C: velocity and acceleration are parallel and both are perpendicular to the displacement. Incorrect, acceleration and displacement are anti-parallel.

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

Memory Aids:
- Acceleration opposes displacement in SHM.

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

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