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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-240 of 483 questions

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211 Medium
211 of 483
Application +4 | -1
In SHM, potential energy of a particle at mean position is E1E_1 and kinetic energy is E2E_2, then
View answer and explanation
Correct option: B
total potential energy at x=3A2x=\frac{\sqrt{3} A}{2} is E1+3E24E_1+\frac{3 E_2}{4}
The correct answer is that the total potential energy at x=3A2x=\frac{\sqrt{3} A}{2} is E1+3E24E_1+\frac{3 E_2}{4} because it correctly applies the potential energy formula for SHM.
212 Medium
212 of 483
Application +4 | -1
A simple pendulum of length 1 m1 \mathrm{~m} and having a bob of mass 100 g100 \mathrm{~g} is suspended in a car. moving on a circular track of radius 100 m100 \mathrm{~m} with uniform speed 10 m/s10 \mathrm{~m} / \mathrm{s} If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by T=2π/α1/4T=2 \pi / \alpha^{1 / 4} The value of α\alpha is

[Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 ]
View answer and explanation
Correct option: C
101
Option C (101) is correct because the time period of the pendulum is affected by the centripetal acceleration of the car, and the effective acceleration is used to calculate the time period.
213 Medium
213 of 483
Analyzing +4 | -1
Consider one dimensional motion of a particle of mass mm. It has potential energy U=a+b2U=a+b^2, where ' aa ' and ' bb ' are positive constants. At origin (x=0)(x=0) it has initial velocity v0v_0. It performs simple harmonic oscillations. The frequency of the simple harmonic motion depends on
View answer and explanation
Correct option: A
bb and mm alone
Option A (bb and mm alone) is correct because the frequency of SHM is derived from the potential energy function and depends only on the constant bb and the mass mm.
214 Medium
214 of 483
Understanding +4 | -1
How does the time period of pendulum vary with length
View answer and explanation
Correct option: A
L\sqrt L
Option A (L\sqrt L) is correct because the time period of a simple pendulum is directly proportional to the square root of its length.
215 Medium
215 of 483
Application +4 | -1
The maximum height of a simple pendulum with respect to the equilibrium position is h\mathrm{h} and the maximum velocity of the bob of pendulum is v\mathrm{v}. Then
View answer and explanation
Correct option: C
v=2gh\mathrm{v}=\sqrt{2 \mathrm{gh}}
Option C (v=2gh\mathrm{v}=\sqrt{2 \mathrm{gh}}) is correct because it is derived from the conservation of energy, equating the change in kinetic energy to the change in potential energy.
216 Medium
216 of 483
Application +4 | -1
In a SHM, when the displacement is one half
the amplitude, what fraction of the total energy
is kinetic?
View answer and explanation
Correct option: D
3/43 / 4
Option D (3/43 / 4) is correct because the kinetic energy is 3/4 of the total energy when the displacement is half the amplitude in SHM.
217 Medium
217 of 483
Application +4 | -1
A simple harmonic oscillation is represented by the equation y=0.40sin⁡(40t7+0.61)y=0.40 \sin \left(\frac{40 t}{7}+0.61\right) where yy and tt are in metre and second respectively. The value of time period is
View answer and explanation
Correct option: A
7π20 sec\frac{7 \pi}{20} \mathrm{~sec}
Option A (7π20 sec\frac{7 \pi}{20} \mathrm{~sec}) is correct because the time period is calculated from the angular frequency in the given equation.
218 Medium
218 of 483
Application +4 | -1
The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cmcm and the mass of the particle is 5.12 kg, the time period of its oscillation is
View answer and explanation
Correct option: A
π5s\frac{\pi }{5}s
Option A (π5s\frac{\pi }{5}s) is correct because it's calculated using the maximum kinetic energy, amplitude, and mass to find the angular frequency and then the time period.
219 Medium
219 of 483
Understanding +4 | -1
For a simple pendulum, the graph between T2T^2 and LL (where TT is the time period and LL is the length) is
View answer and explanation
Correct option: A
straight line passing through origin
Option A (straight line passing through origin) is correct because the square of the time period (T2T^2) is directly proportional to the length (LL) of the pendulum, resulting in a linear relationship that passes through the origin.
220 Medium
220 of 483
Understanding +4 | -1
If <E><{E}> and <U>< {U} > denote the average kinetic and the average potential energies respectively of mass describing a simple harmonic motion, over one period, then the correct relation is
View answer and explanation
Correct option: A
<E>=<U>< E > = < U >
Option A is correct because in Simple Harmonic Motion (SHM), the average kinetic energy and average potential energy over one period are equal.
221 Easy
221 of 483
Analyzing +4 | -1
Molten-wax of mass mm drops on a block of mass MM, which is oscillating on a frictionless table. Select the correct option.
View answer and explanation
Correct option: D
time period increases
Option D (time period increases) is correct because the time period of a simple harmonic oscillator is proportional to the square root of the mass, and the mass increases when wax drops on the block.
222 Medium
222 of 483
Application +4 | -1
The displacement of an elastic wave is given by the function y=3sin⁡ωt+4cos⁡ωt,y = 3\sin \omega t + 4\cos \omega t, where ‘y’ is in cm and ‘t’ is in second. Then the resultant amplitude in
View answer and explanation
Correct option: D
5
Option D (5) is correct because the resultant amplitude is found using the formula for the superposition of two waves with a phase difference of 90 degrees.
223 Medium
223 of 483
Application +4 | -1
A linear oscillator of unit mass of force constant 4×106  Nm4 \times {10^6}\,\,\frac{N}{m} and amplitude 0.01   m0.01\,\,\,m has a total mechanical energy of
View answer and explanation
Correct option: A
200  J200\,\,J
Option B (6 m) is correct because the length of the rod is calculated from the radius of gyration formula.
224 Medium
224 of 483
Application +4 | -1
A particle hanging from a spring stretches it by 1 cm1 \mathrm{~cm} at earth's surface. Radius of earth is 6400 km6400 \mathrm{~km}. At a place 800 km\mathrm{km} above the earth's surface, the same particle will stretch the spring by:
View answer and explanation
Correct option: D
0.79 cm
Option D (0.79 cm) is correct because it is calculated by considering the variation of gravitational force with distance from the Earth's center.
225 Medium
225 of 483
Understanding +4 | -1
Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
View answer and explanation
Correct option: A
Simple harmonic motion
Option A (Simple harmonic motion) is correct because the ball bearing's motion approximates SHM when released slightly above the lowest point in a smooth curved bowl.
226 Medium
226 of 483
Understanding +4 | -1
The total energy of particle performing SHM depend on
View answer and explanation
Correct option: A
k,a,mk, a, m
Option A (k,a,mk, a, m) is correct because the total energy of a particle in SHM depends on the spring constant (kk), amplitude (aa), and mass (mm).
227 Easy
227 of 483
Understanding +4 | -1
Two particles are executing simple harmonic motion. At an instant of time tt, their displacements are y1=Acos⁡(ωt)y_1=A \cos (\omega t) and y2=Asin⁡(ωt)y_2=A \sin (\omega t)
Then, the phase difference between y1y_1 and y2y_2 is
View answer and explanation
Correct option: B
90∘90^{\circ}
Option B (90∘90^{\circ}) is correct because the phase difference between cosine and sine functions with the same argument is 90 degrees.
228 Medium
228 of 483
Application +4 | -1
A particle is executing S.H.M with time period TT and amplitude AA. The graphical plot of velocity versus displacement (v−x)(v-x) of the particle
View answer and explanation
Correct option: C
May be a circle
Option C (May be a circle) is correct because the v-x plot for SHM can be an ellipse, and a circle is a special case of an ellipse when the angular frequency is 1.
229 Medium
229 of 483
Application +4 | -1
A particle is executing simple harmonic motion. Its displacement to amplitude ratio when its kinetic energy is 84%84 \% of total energy is
View answer and explanation
Correct option: B
2:52: 5
Option B (2:52:5) is correct because the ratio is found by equating the kinetic energy to 84% of the total energy and solving for the displacement to amplitude ratio.
230 Medium
230 of 483
Application +4 | -1
A body is executing SHM, if the force acting on the body is 6 N. When the displacement is 2 cm, then the force acting on the body at a displacement of 3 cm is
View answer and explanation
Correct option: B
9N
Option B (9N) is correct because the force in SHM is directly proportional to the displacement.
231 Medium
231 of 483
Application +4 | -1
Three springs are connected to a mass mm. When mass mm oscillates, what is the effective spring constant and time period of vibration? Given k=2 Nm−1k=2 \mathrm{~Nm}^{-1} and m=80 gm=80 \mathrm{~g}.
View answer and explanation
Correct option: A
8 Nm−1,0.628 s8 \mathrm{~Nm}^{-1}, 0.628 \mathrm{~s}
The correct answer is A (8 Nm−1,0.628 s8 \mathrm{~Nm}^{-1}, 0.628 \mathrm{~s}) because the effective spring constant is the sum of individual spring constants and the time period is calculated using this effective constant.
232 Easy
232 of 483
Application +4 | -1
To make the frequency of an oscillator double, one has to
View answer and explanation
Correct option: D
reduce the mass to one-fourth
Option D (reduce the mass to one-fourth) is correct because the frequency of an oscillator is inversely proportional to the square root of the mass.
233 Medium
233 of 483
Application +4 | -1
If a spring of mass 30 kg30 \mathrm{~kg} has spring constant of 15 N/m15 \mathrm{~N} / \mathrm{m}, then its time period, is
View answer and explanation
Correct option: D
2π22 \pi \sqrt{2} seconds
Option D (2π22 \pi \sqrt{2} seconds) is correct because the time period of a spring is calculated using the formula T=2πmkT=2 \pi \sqrt{\frac{m}{k}}.
234 Medium
234 of 483
Understanding +4 | -1
Statement-I: If Amplitude of SHM is doubled, the periodicity wall remain same.

Statement-II: Amlitude and periodicity are two independent characteristics of SHM.
View answer and explanation
Correct option: A
If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement-I.
Option A (If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement-I) is correct because the period of SHM is independent of the amplitude.
235 Medium
235 of 483
Analyzing +4 | -1
A pendulum made of a uniform wire of cross – sectional area A has time period T. When an additional mass M is added to its bob, the time period changes TM.{T_M}. If the Young’s modulus of the material of the wire is Y, then 1Y\frac{1}{Y} is equal to ( gg = gravitational acceleration)
View answer and explanation
Correct option: A
[(TMT)2−1]AMg\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}}
Option A ([(TMT)2−1]AMg\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{Mg}}) is correct because it correctly relates the change in time period to Young's modulus and the added mass.
236 Medium
236 of 483
Application +4 | -1
A particle is executing SHM of period 24 xx and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q. where OP = -9 cm and OQ =40 cm is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the minimum time taken to go from P to Q is calculated using the SHM equation and the given displacements.
237 Medium
237 of 483
Application +4 | -1
The displacement of an elastic wave is given by the function y=3sin⁡ωt+4cos⁡ωt,y = 3\sin \omega t + 4\cos \omega t, where ‘y’ is in cm and ‘t’ is in second. Then the resultant amplitude in
View answer and explanation
Correct option: D
5
Option D (5) is correct because the resultant amplitude is found using the formula for the superposition of two waves with a phase difference of 90 degrees.
238 Medium
238 of 483
Understanding +4 | -1
Which one of the following statements is
correct?
View answer and explanation
Correct option: C
The measurement of mass taken by a spring weighing balance is corrected at the place where the acceleration due to gravity is same with the place where the spring balance is calibrated for
The measurement of mass taken by a spring weighing balance is corrected at the place where the acceleration due to gravity is same with the place where the spring balance is calibrated for is correct because the reading depends on the local gravitational acceleration.
239 Medium
239 of 483
Application +4 | -1
At a given instant of time the position vector of a particle moving in a circle with a velocity 3i^−4j^+5k^3 \hat{i}-4 \hat{j}+5 \hat{k} is i^+9j^−8k^\hat{i}+9 \hat{j}-8 \hat{k}. Its angular velocity at that time is
View answer and explanation
Correct option: B
(13i^−29j^−31k^)146\frac{(13 \hat{i}-29 \hat{j}-31 \hat{k})}{146}
Option D ((13i^−29j^−31k^)146\frac{(13 \hat{i}-29 \hat{j}-31 \hat{k})}{146}) is correct because it is calculated using the formula for angular velocity, which is the cross product of the position vector and velocity vector, divided by the square of the magnitude of the position vector.
240 Medium
240 of 483
Understanding +4 | -1
When a simple pendulum is moved from the Earth's surface to deep mine, the period of oscillation
View answer and explanation
Correct option: C
Increases
Option C is correct because the acceleration due to gravity decreases as you go deeper into the mine, which increases the period of oscillation of a simple pendulum.
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