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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Medium

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Practice format and chapter quiz

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121 Medium
121 of 445
Analyzing +4 | -1
Assertion: We can call circular motion also as simple harmonic motion.

Reason: Angular velocity in uniform circular motion and angular frequency in simple harmonic motion have the same meanings.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true
Option D (If Assertion is false but the Reason is true) is correct because circular motion is not simple harmonic motion, but the angular velocity in uniform circular motion and angular frequency in simple harmonic motion have related meanings.
122 Medium
122 of 445
Application +4 | -1
For definite length of wire, if the weight used for applying tension is immersed in water, then frequency will
View answer and explanation
Correct option: A
Become less
Option A (Become less) is correct because when the weight is immersed in water, the effective tension decreases due to buoyancy, thus reducing the frequency.
123 Medium
123 of 445
Analyzing +4 | -1
The displacement of a particle executing periodic motion is given by y=4cos⁡2(t/2)sin⁡(1000t)y=4 \cos ^2(t / 2) \sin (1000 t). Find the independent constituent SHM's.
View answer and explanation
Correct option: A
sin⁡(1001t),sin⁡(1000t),sin⁡(999t)\sin (1001 t), \sin (1000 t), \sin (999 t)
Option A (sin⁡(1001t),sin⁡(1000t),sin⁡(999t)\sin (1001 t), \sin (1000 t), \sin (999 t)) is correct because the given equation can be simplified using trigonometric identities to a sum of these three sine functions.
124 Medium
124 of 445
Understanding +4 | -1
Energy of particle executing SHM depends upon
View answer and explanation
Correct option: B
Amplitude and frequency
Option B is correct because the energy of a particle in SHM depends on both the amplitude and the frequency, as shown in the formula E=12mω2A2E = \frac{{1}}{{2}} m \omega^2 A^2.
125 Medium
125 of 445
Analyzing +4 | -1
A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm40 {~cm}. The block just loses contact with the plank when the later is momentarily at rest, then
View answer and explanation
Correct option: B
The block weighs double its weight when the plank is at one of the positions of momentary at rest
Option B is correct because at the extreme position, the block's weight equals the restoring force, and at the other extreme, both act downward, doubling the effective weight.
126 Medium
126 of 445
Understanding +4 | -1
The system that returns to equilibrium as quickly as possible without oscillating is
View answer and explanation
Correct option: B
Critically damped
Option B (Critically damped) is correct because a critically damped system returns to equilibrium the fastest without oscillating.
127 Medium
127 of 445
Remembering +4 | -1
x(t)=Acos⁡(ωt+ϕ)x(t)=A \cos (\omega t+\phi) is the equation of simple harmonic motion. In this equation ' ϕ\phi ' is called
View answer and explanation
Correct option: A
phase constant
Option A (phase constant) is correct because 'ϕ\phi' in the equation of simple harmonic motion represents the phase constant.
128 Medium
128 of 445
Analyzing +4 | -1
S-I: If the amplitude of the oscillation is doubled the maximum kinetic energy becomes four times.
S-II: If the amplitude of the oscillation is doubled the minimum potential energy also becomes four times.
View answer and explanation
Correct option: C
S-I is true &\& S-II is false
Option C is correct because S-I is true and S-II is false. Maximum kinetic energy is proportional to the square of the amplitude, while minimum potential energy is zero at the mean position and independent of amplitude.
129 Medium
129 of 445
Application +4 | -1
Instantaneous acceleration (in ms−2{ms}^{-2} ) of a particle executing S.H.M is given by a=−π4sin⁡(π4t−π6)a=-\frac{\pi}{4} \sin \left(\frac{\pi}{4} t-\frac{\pi}{6}\right). The maximum speed of the particle will occur first time at
View answer and explanation
Correct option: D
0.67 s0.67 {~s}
Option D (0.67 {~s}) is correct because maximum speed occurs at the mean position where acceleration is zero, which can be found by setting the acceleration equation to zero and solving for t.
130 Medium
130 of 445
Application +4 | -1
A particle starts executing simple harmonic motion (SHM) of amplitude aa and total energy EE. At any instant, its kinetic energy is 3E4\frac{{3E}}{4}, then its displacement yy is given by
View answer and explanation
Correct option: D
y=a2y = \frac{a}{2}
Option D (y=a2y = \frac{a}{2}) is correct because it is derived from the relationship between kinetic energy, total energy, amplitude, and displacement in SHM.
131 Medium
131 of 445
Understanding +4 | -1
Resonance is an example of:
View answer and explanation
Correct option: B
Forced vibration
Option B (Forced vibration) is correct because resonance is a phenomenon that occurs when a system is driven by an external force at its natural frequency, which is a type of forced vibration.
132 Medium
132 of 445
Application +4 | -1
A particle executes linear S.H.M. with amplitude 4 cm4 \mathrm{~cm}. The magnitude of velocity and acceleration is equal when it is at 3 cm3 \mathrm{~cm} from mean position. The time period is
View answer and explanation
Correct option: C
6π7\frac{6 \pi}{\sqrt{7}}
Option C (6π7\frac{6 \pi}{\sqrt{7}}) is correct because it is calculated from the condition where the magnitude of velocity equals the magnitude of acceleration in SHM.
133 Medium
133 of 445
Application +4 | -1
The period of a simple pendulum inside a stationary lift isTT. The lift accelerates upwards with an acceleration of g⁄3. The time period of pendulum will be
View answer and explanation
Correct option: C
32T\frac{{\sqrt 3 }}{2}T
Option C (32T\frac{{\sqrt 3 }}{2}T) is correct because the effective acceleration due to the upward acceleration of the lift increases the apparent weight, affecting the pendulum's period.
134 Medium
134 of 445
Analyzing +4 | -1
Assertion (A): The graph of PE and KE of a particle is SHM with respect to position is a parabola.
Reason (R): This because PE and KE do not vary linearly with position.
View answer and explanation
Correct option: B
If Both A &\& R are true but R is not correct explanation of A
Option B is correct because both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion. The parabolic nature is due to the quadratic relationship with position.
135 Medium
135 of 445
Understanding +4 | -1
The time period of a simple pendulum on a freely moving artificial satellite is
View answer and explanation
Correct option: D
infinite
Option D is correct because, in a freely moving artificial satellite, the effective gravitational acceleration is zero, leading to an infinite time period for a simple pendulum.
136 Medium
136 of 445
Analyzing +4 | -1
Which of the following quantities are always negative in a S.H.M
(F→= force,  V→ = velocity,  a→ = acceleration,  r→ = displacement)(\overrightarrow F = \,force,\,\,\overrightarrow V \, = \,velocity,\,\,\overrightarrow a \, = \,acceleration,\,\,\overrightarrow r \, = \,displacement)
View answer and explanation
Correct option: B
a→ ⋅r→\overrightarrow a \, \cdot \overrightarrow r
Option B is correct because in SHM, acceleration and displacement are always in opposite directions, making their dot product always negative.
137 Medium
137 of 445
Understanding +4 | -1
Assertion: The graph of total energy of a particle in SHM. wrt, position is a straight line with zero slope.
Reason: Total energy of particle in SHM remains constant throughout its motion.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion
Option A is correct because the total energy in SHM remains constant, therefore the graph is a horizontal line.
138 Medium
138 of 445
Application +4 | -1
A particle of mass 4 cm4 {~cm} is executing oscillations about the origin on the XX-axis. Its potential energy is U(x)=k∣x∣3U(x)=k|x|^3, where kk is a positive constant. If the amplitude of oscillation is 4 cm4 {~cm}, then its time period TT will be proportional to
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because the time period T is proportional to a(-0.5) based on the given potential energy function.
139 Medium
139 of 445
Understanding +4 | -1
Particle moves on the x\mathrm{x}-axis according to the equation x=\mathrm{x}= A+Bsin⁡ωt\mathrm{A}+\mathrm{B} \sin \omega \mathrm{t}. The motion is simple harmonic with amplitude.
View answer and explanation
Correct option: B
B
Option B (B) is correct because in the equation x = A + B sin(ωt), B represents the amplitude of the sinusoidal motion.
140 Medium
140 of 445
Application +4 | -1
How much revolution does the engine make during the time when a motor wheel with angular speed is increased from 720 rpm720 \mathrm{~rpm} to 2820 rpm2820 \mathrm{~rpm} in 14 seconds?
View answer and explanation
Correct option: C
413
Option C (413) is correct because it is calculated by finding the angular acceleration and then the total angular displacement, finally dividing by 2π2\pi to get the number of revolutions.
141 Medium
141 of 445
Application +4 | -1
An object of mass 0.2 kg0.2 \mathrm{~kg} executes SHM\mathrm{SHM} along xx-axis with frequency of 25π\frac{25}{\pi} Hz. At the position x=0.04 mx=0.04 \mathrm{~m}, the object has kinetic energy of 0.5 J0.5 \mathrm{~J} and potential energy of 0.4 J0.4 \mathrm{~J}. The amplitude of oscillation is
View answer and explanation
Correct option: B
0.06 m0.06 \mathrm{~m}
Option B (0.06 m0.06 \mathrm{~m}) is correct because it is calculated using the total energy and the angular frequency of the SHM.
142 Medium
142 of 445
Understanding +4 | -1
The amplitude of vibration of a particle is given by am=(a0)/(aω2−bω+c);{a_m} = ({a_0})/(a{\omega ^2} - b\omega + c); where a0,a,b{a_0},a,b and c are positive. The condition for a single resonant frequency is
View answer and explanation
Correct option: A
b2=4ac{b^2} = 4ac
Option A (b2=4ac{b^2} = 4ac) is correct because for a single resonant frequency, the denominator of the amplitude equation must have only one real root, which occurs when the discriminant is zero.
143 Medium
143 of 445
Application +4 | -1
The height of liquid column in a UU tube is 0.3 meter. If the liquid in one of the limbs is depressed and then released, then the time period of liquid column will be-
View answer and explanation
Correct option: C
1.1 sec1.1 \mathrm{~sec}
Option C (1.1 sec) is correct because it is the calculated time period of oscillation for the liquid column in the U-tube.
144 Medium
144 of 445
Application +4 | -1
If the ratio of the acceleration due to gravity on the surface of earth to that on the surface of the moon is 6:16: 1, then the ratio of the periods of a simple pendulum on their surface is
View answer and explanation
Correct option: D
1:61: \sqrt{6}
Option D (1:61: \sqrt{6}) is correct because the period of a simple pendulum is inversely proportional to the square root of gravity.
145 Medium
145 of 445
Application +4 | -1
If two springs AA and BB with spring constant 2 k2 \mathrm{~k} and k\mathrm{k}, are stretched separately by same suspended weight, then the ratio between the work done in stretching AA and BB is
View answer and explanation
Correct option: A
1:21: 2
Option A (1:21: 2) is correct because the ratio of work done is inversely proportional to the spring constant when stretched by the same weight.
146 Medium
146 of 445
Application +4 | -1
The phase difference between two SHMsSHMs y1=10sin⁡(20t+π6)y_1=10 \sin \left(20 t+\frac{\pi}{6}\right) and y2=20cos⁡(20t−π6)y_2=20 \cos \left(20 t-\frac{\pi}{6}\right) is
View answer and explanation
Correct option: A
π6\frac{\pi}{6}
Option A (25 rev/s2 \sqrt{5} \mathrm{~rev} / \mathrm{s}) is correct because the angular velocity is calculated using the kinematic equation for rotational motion.
147 Medium
147 of 445
Application +4 | -1
A horizontal platform vibrates with simple harmonic motion in the horizontal direction with a period 2 s. A body of mass 0.5 kg is placed on the platform. The coefficient of static friction between the body and platform is 0.3. What is the maximum frictional force on the body when the platform is oscillating with amplitude 0.2 m? Assume π2=10=g{\pi ^2} = 10 = g.
View answer and explanation
Correct option: B
1 N
Option B (1 N) is correct because the maximum frictional force is limited by the maximum force on the body due to SHM, which is less than the maximum static friction.
148 Medium
148 of 445
Application +4 | -1
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes (14)th \left(\frac{1}{4}\right)^{\text {th }} of the original in time ' tt ' and ' nn ' revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform)
View answer and explanation
Correct option: C
16n15\frac{16 n}{15}
Option C (16n15\frac{16 n}{15}) is correct because it is derived from the equations of rotational motion with uniform angular retardation.
149 Medium
149 of 445
Application +4 | -1
A body is executing simple harmonic motion. When the displacements from the mean position is 4 cm4 {~cm} and 5 cm5 {~cm}, the corresponding velocities of the body are 10 cm/sec10 {~cm} / {sec} and 8 cm/sec8 {~cm} / {sec}. Then the time period of the body is
View answer and explanation
Correct option: C
π sec\pi\, {sec}
Option C (π sec\pi \text{{ sec}}) is correct because it is the calculated time period using the given velocities and displacements in simple harmonic motion.
150 Medium
150 of 445
Analyzing +4 | -1
The total mechanical energy of a spring-mass system in simple harmonic motion is E=12mω2A2E=\frac{1}{2} m \omega^2 A^2. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude AA remains the same. The new mechanical energy will
View answer and explanation
Correct option: D
remain EE
Option D (remain EE) is correct because the total mechanical energy depends on the spring constant and amplitude, not the mass, when expressed in terms of kk and AA.
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