When two displacements represented by y 1 =asin(ωt) and y 2 =bcos(ωt) are superimposed,the motion is -

simple harmonic with amplitude a 2 + b 2

simple harmonic with amplitude a/b

simple harmonic with amplitude (a+b)/2

Physics - Oscillations - Quiz-1

The motion of a particle varies with time according to the relation $y = a \sin ωt + a \cos ωt$ . Then

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The motion is oscillatory but not SHM
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The motion is SHM with an amplitude $a \sqrt{2}$
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The motion is SHM with an amplitude $\sqrt{2}$
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The motion is SHM with an amplitude a

If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its time period are, respectively,

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2A, A
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4A, 0
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A, A
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0, 2A

The variation of acceleration of a particle executing SHM with displacement x is:

A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them will be

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$\frac{π}{3}$
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$\frac{2 π}{3}$
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$\frac{π}{6}$
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$\frac{π}{2}$

The equations of the displacement of two particles making SHM are represented by y₁ = a sin (ωt + φ) and y₂ = a cos (ωt) respectively. The phase difference of the velocities of the two particles will be

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π/2 + φ
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-φ
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φ
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φ - π/2

The displacement of a particle executing SHM is given by y = 0.25 (sin 200t) cm. The maximum speed of the particles is:

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200 cm/sec
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100 cm/sec
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50 cm/sec
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0.25 cm/sec

Which of the following figure represents damped harmonic motion?
(i)
(ii)
(iii)
(iv)

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(i) and (ii)
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(iii) and (iv)
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(i), (ii), (iii), and (iv)
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(i) and (iv)

A particle is executing SHM with an amplitude A and the time period T. If at t = 0, the particle is at origin (mean position), then the time instant when it covers a distance equal to 2.5A will be

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$\frac{T}{12}$
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$\frac{5 T}{12}$
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$\frac{7 T}{12}$
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$\frac{2 T}{3}$

The time period of a spring mass system at the surface of earth is 2 second. What will be the time period of this system on the moon where acceleration due to gravity is $\frac{1}{6} th$ of the value of g on earth's surface?

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$\frac{1}{\sqrt{6}} seconds$
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$2 \sqrt{6} seconds$
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$2 seconds$
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$12 seconds$

A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

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$\frac{1}{2} s$
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$\frac{1}{6} s$
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$\frac{1}{4} s$
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$\frac{1}{3} s$

The uniform stick of mass m length L is pivoted at the centre. In the equilibrium position shown in the figure, the identical light springs have their natural length. If the stick is turned through a small angle $θ$ , it executes SHM. The frequency of the motion is:

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$\frac{1}{2 π} \sqrt{\frac{6 K}{m}}$
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$\frac{1}{2 π} \sqrt{\frac{3 K}{2 m}}$
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$\frac{1}{2 π} \sqrt{\frac{3 K}{m}}$
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None of these

If the displacement x and the velocity v of a particle executing simple harmonic motion are related through the expression $4 v^{2} = 25 - x^{2}$ ,then its time period will be

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$π$
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2 $π$
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4 $π$
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6 $π$

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $\vec{a}$ is correctly shown in

1.
2.
3.
4.

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1
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2
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3
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4

A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket:

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Comes down with uniform acceleration
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Moves around the earth in a geostationary orbit
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Moves up with a uniform velocity
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Moves up with the uniform acceleration

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is

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0.8 T
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0.25 T
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2 T
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4 T

A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. μ_s is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q will be

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kA
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$\frac{k A}{2}$
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Zero
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μ_s mg

A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is T₀. When parallel plates are charged, the electric field between the plates is E and time period changes to T. The ratio T/T₀ is equal to

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${(\frac{g + \frac{q E}{m}}{g})}^{1 / 2}$
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${(\frac{g}{g + \frac{q E}{m}})}^{3 / 2}$
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${(\frac{g}{g + \frac{q E}{m}})}^{1 / 2}$
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None of these

A simple pendulum has a time period $T_{1}$ when on the earth’s surface, and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $T_{2} / T_{1}$ is:

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1
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$\sqrt{2}$
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4
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2

A particle executes linear simple harmonic motion with an amplitude of of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is

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() $\sqrt{\frac{5}{π}}$
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() $\sqrt{\frac{5}{2 π}}$
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() $\frac{4 π}{\sqrt{5}}$
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() $\frac{2 π}{\sqrt{3}}$

A body mass m is attached to the lower end of a spring whose upper end is fixed. The spring has neglible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. The value of m in kg is-

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$\frac{3}{4}$
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$\frac{4}{3}$
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$\frac{16}{9}$
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$\frac{9}{16}$

When two displacements represented by y₁=asin(ωt) and y₂=bcos(ωt) are superimposed,the motion is -

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not a simple harmonic
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simple harmonic with amplitude a/b
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simple harmonic with amplitude $\sqrt{a^{2} + b^{2}}$
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simple harmonic with amplitude (a+b)/2

The damping force on an oscillator is directly proportional to the velocity.The units of the constant of proportionality are

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$k g m s^{- 1}$
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$k g m s^{- 2}$
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$k g s^{- 1}$
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$k g s$

The displacement of a particle along the x-axis is given by $x = a \sin^{2} ω t$ . The motion of the particle corresponds to:

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simple harmonic motion of frequency $ω / π.$
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simple harmonic motion of frequency $3 ω / 2 π.$
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non-simple harmonic motion.
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simple harmonic motion of frequency $ω / 2 π.$

The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

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T
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T/ $\sqrt{2}$
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2T
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$\sqrt{2} T$

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

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0.5 $π$
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$π$
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0.707 $π$
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zero

A body performs simple harmonic motion about x=0 with an amplitude a and a time period T. The speed of the body at $x = \frac{a}{2}$ will be:

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$\frac{πa \sqrt{3}}{2 T}$
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$\frac{πa}{T}$
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$\frac{3 π^{2} a}{T}$
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$\frac{πa \sqrt{3}}{T}$

Which one of the following equations of motion represents simple harmonic motion?

where k, $k_{0}, k_{1}$ and $α$ are all positive.

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Acceleration =- $k_{0} x + k_{1} x^{2}$
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Acceleration =- $k (x + a)$
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Acceleration =k $(x + a)$
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Acceleration =kx

Two simple harmonic motions of angular frequency 100 rad s ^-1 and 1000 rad $s^{- 1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be

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1:10
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1:10²
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1:10³
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1:10⁴

A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin $(ω t + π / 6)$ .After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?

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$\frac{T}{8}$
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$\frac{T}{6}$
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$\frac{T}{3}$
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$\frac{T}{12}$

An S.H.M. has an amplitude ‘a’ and a time period T. The maximum velocity will be -

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${4a \over T}$
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${2a \over T}$
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${2 \pi \over T}$
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${2a \pi \over T}$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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