A particle is executing SHM along a straight line. Its velocities at distances x 1 and x 2 from the mean position are v 1 and v 2 , respectively. Its time period is:

$2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}}{\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}}}$

$2 \pi \sqrt{\frac{x_{1}^{2}+x_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}~$

$2 \pi \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{x_{1}^{2}+x_{2}^{2}}}$

$2 \pi \sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}}$

Physics - Oscillations - Quiz-3

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

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$T = 2 π \sqrt{\frac{R_{e}}{g}}$
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$T = 2 π \sqrt{\frac{2 R_{e}}{g}}$
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$T = 2 π \sqrt{\frac{R_{e}}{2 g}}$
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T=2 seconds

If the displacement equation of a particle be represented by $y = A \sin P t + B \cos P t$ , the particle executes

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A uniform circular motion
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A uniform elliptical motion
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A S.H.M.
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A rectilinear motion

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $Fsinωt$ . If the amplitude of the particle is maximum for $ω = ω_{1}$ and the energy of the particle is maximum for $ω = ω_{2}$ , then (where $ω_{0}$ is natural frequency of oscillation of particle)

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$ω_{1} = ω_{0}$ and $ω_{2} \neq ω_{0}$
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$ω_{1} = ω_{0}$ and $ω_{2} = ω_{0}$
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$ω_{1} \neq ω_{0}$ and $ω_{2} = ω_{0}$
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$ω_{1} \neq ω_{0}$ and $ω_{2} \neq ω_{0}$

The displacement of a particle varies according to the relation $x = 4 (cosπ t + sinπ t) .$ The amplitude of the particle is

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8
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– 4
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4
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$4 \sqrt{2}$

A S.H.M. is represented by $x = 5 \sqrt{2} (\sin 2 πt + \cos 2 πt) .$ The amplitude of the S.H.M. is

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10 cm
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20 cm
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$5 \sqrt{2}$ cm
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50 cm

Amplitude of a wave is represented by

$A = \frac{c}{a + b - c}$

Then resonance will occur when

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$b = - c / 2$
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b = 0 and a = c
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$b = - a / 2$
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None of these

The displacement of a particle varies with time as $x = 12 \sin w t - 16 \sin^{3} w t$ (in cm). If its motion is S.H.M., then its maximum acceleration is -

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() $12 ω^{2}$
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() $36 ω^{2}$
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() $144 ω^{2}$
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() $\sqrt{192} ω^{2}$

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3}$ , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is -

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Proportional to $\frac{1}{\sqrt{a}}$
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Independent of a
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Proportional to $\sqrt{a}$

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Proportional to $a^{3 / 2}$

A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be

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$T = 2 π \sqrt{(\frac{M h}{P A})}$
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$T = 2 π \sqrt{(\frac{M A}{P h})}$
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$T = 2 π \sqrt{(\frac{M}{P A h})}$
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$T = 2 π \sqrt{M P h A}$

The metallic bob of a simple pendulum has the relative density $ρ$ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

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$T \frac{ρ - 1}{ρ}$
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$T \frac{ρ}{ρ - 1}$
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$T \sqrt{\frac{ρ - 1}{ρ}}$
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$T \sqrt{\frac{ρ}{ρ - 1}}$

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire is A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to -

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$2 π (\frac{m}{K})$
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$2 π {\{\frac{(YA + KL) m}{YAK}\}}^{1 / 2}$
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$2 π \frac{mYA}{KL}$
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$2 π \frac{mL}{YA}$

On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as massless) is

(a) $2 π {(\frac{M}{2 K})}^{1 / 2}$ (b) $2 π {(\frac{2 M}{K})}^{1 / 2}$

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1
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2
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3
0%
4

An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is -

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4 Mg/K
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2 Mg/K
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Mg/K
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Mg/2K

A particle of mass m is attached to three identical springs A, B and C each of force constant k a shown in figure. If the particle of mass m is pushed slightly against the spring A and released then the time period of oscillations is -

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() $2 π \sqrt{\frac{2 m}{k}}$
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() $2 π \sqrt{\frac{m}{2 k}}$
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() $2 π \sqrt{\frac{m}{k}}$
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() $2 π \sqrt{\frac{m}{3 k}}$

The graph shows the variation of displacement of a particle executing S.H.M. with time. We infer from this graph that -

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The force is zero at time T/8
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The velocity is maximum at time T/4
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The acceleration is maximum at time T
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The P.E. is equal to total energy at time T/4

For a particle executing S.H.M. the displacement x is given by $A \cos ω t$ . Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x.

(a) I, III (b) II, IV

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1
0%
2
0%
3
0%
4

The velocity-time diagram of a harmonic oscillator is shown in the adjoining figure. The frequency of oscillation is

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25 Hz
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50 Hz
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12.25 Hz
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33.3 Hz

The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

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1 $\times$ 10² N/m
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150 N/m
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0.667 $\times$ 10² N/m
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3 $\times$ 10² N/m

A body performs S.H.M. . Its kinetic energy K varies with time t as indicated by graph

(a) (b)

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1
0%
2
0%
3
0%
4

The displacement of a body executing SHM is given by x= A sin (2 $π$ t + $π$ /3). The first time from t = 0 when the velocity is maximum is :

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0.33 sec
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0.16 sec
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0.25 sec
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0.5 sec

The time period of a particle executing SHM is 8 sec. At t = 0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is :

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$\frac{1}{\sqrt{2} + 1}$
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$\sqrt{2}$
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$\frac{1}{\sqrt{2}}$
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$\sqrt{2} + 1$

Two particles are in SHM in a straight line. Amplitude A and time period T of both the particles are equal. At time t = 0, one particle is at displacement Y₁ = + A and the other at Y₂ = -A/2, and they are approaching towards each other. After what time they cross each other?

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T/3
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T/4
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5T/6
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T/6

Two particles execute SHM of the same amplitude of 20 cm with the same time period along the same line about the same equilibrium position. The maximum distance between the two is 20 cm. Their phase difference in radians is :

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$\frac{2 π}{3}$
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$\frac{π}{2}$
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$\frac{π}{3}$
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$\frac{π}{4}$

Two particles P and Q describe simple harmonic motions of the same period, the same amplitude, along the same line about the same equilibrium position O. When P and Q are on opposite sides of O at the same distance from O, they have the same speed of 1.2 m/s in the same direction. When their displacements are the same, they have the same speed of 1.6 m/s in opposite directions. The maximum velocity in m/s of either particle is

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2.8
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2.5
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2.4
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2

A pendulum consisting of a small sphere of mass m suspended by an inextensible and massless string of length l is made to swing in a vertical plane. If the breaking strength of the string is 2mg, then the maximum angular amplitude of the displacement from the vertical can be :-

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0 $°$
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30 $°$
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60 $°$
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90 $°$

A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is k^'. Then they are connected in parallel and force constant is k’’. Then k’: k’’ is:

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1:9
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1:11
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1:14
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1:6

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is:

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2 $π$ s
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$π$ s
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2 s
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1 s

A particle is executing a simple harmonic motion. Its maximum acceleration is $α$ and maximum velocity is $β$ . Then its time period of vibration will be:

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$\frac{β^{2}}{α^{2}}$
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$\frac{β}{α}$
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$\frac{β^{2}}{α}$
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$\frac{2 π β}{α}$

A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are v₁ and v₂, respectively. Its time period is:

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$2 \pi \sqrt{\frac{x_{1}^{2}+x_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}~$
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$2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}}{\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}}}$
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$2 \pi \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{x_{1}^{2}+x_{2}^{2}}}$
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$2 \pi \sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}}$

The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos (ωt)

where X = displacement at time t
ω = frequency of oscillation
Which one of the following graphs shows correctly the variation 'a' with 't'?
Here a = acceleration at time t
T = time period

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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