A pendulum has time period \(T\). If it is taken on to another planet having acceleration due to gravity half and mass \(9\) times that of the earth, then its time period on the other planet will be:
1. \(\sqrt{T} \) 2. \(T \)
3. \({T}^{1 / 3} \) 4. \(\sqrt{2} {T}\)

Subtopic:  Angular SHM |
 83%
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A particle in SHM is described by the displacement equation xt=Acos ωt+θ. If the initial position of the particle is 1 cm and its initial velocity is πcm/s, what is its amplitude? (The angular frequency of the particle is π s-1)

(1)   1 cm       

(2)  2 cm

(3)    2 cm     

(4)   2.5 cm

Subtopic:  Simple Harmonic Motion |
 58%
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A simple pendulum hanging from the ceiling of a stationary lift has a time period \(T_1\). When the lift moves downward with constant velocity, then the time period becomes \(T_2\). It can be concluded that: 
1. \(T_2 ~\text{is infinity} \) 2. \(T_2>T_1 \)
3. \(T_2<T_1 \) 4. \(T_2=T_1\)
Subtopic:  Angular SHM |
 62%
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If the length of a pendulum is made \(9\) times and mass of the bob is made \(4\) times, then the value of time period will become:
1. \(3T\)
2. \(\dfrac{3}{2}T\)
3. \(4T\)
4. \(2T\)

Subtopic:  Angular SHM |
 83%
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A simple harmonic wave having an amplitude a and time period T is represented by the equation y=5 sinπt+4m Then the value of amplitude (a) in (m) and time period  (T) in second are       

(1)   a=10, T=2   

(2) a=5, T=1

(3)    a=10, T=1    

(4) a=5, T=2

Subtopic:  Simple Harmonic Motion |
 85%
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The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

(1) T3

(2) T3

(3) 32T

(4) 3T

Subtopic:  Simple Harmonic Motion |
 87%
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The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g3 is

(1) 2π3Lg

(2) π3Lg

(3) 2π3L2g

(4) 2π2L3g

Subtopic:  Simple Harmonic Motion |
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If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period:
1. \(T = 2\pi \sqrt{\frac{R_e}{g}}\)
2. \(T = 2\pi \sqrt{\frac{2R_e}{g}}\)
3. \(T = 2\pi \sqrt{\frac{R_e}{2g}}\)
4. \(T = 2~\text{s}\)
Subtopic:  Simple Harmonic Motion |
 69%
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If the displacement equation of a particle be represented by y=AsinPt+ Bcos Pt , the particle executes

(1)         A uniform circular motion

(2)         A uniform elliptical motion

(3)         A S.H.M.

(4)         A rectilinear motion

Subtopic:  Simple Harmonic Motion |
 82%
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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=ω1  and the energy of the particle is maximum for ω=ω2, then (where ω0 is natural frequency of oscillation of particle)

1. ω1=ω0 and ω2ω0

2. ω1=ω0 and ω2=ω0

3. ω1ω0 and ω2=ω0

4. ω1ω0 and ω2ω0

Subtopic:  Forced Oscillations (OLD NCERT) |
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