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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The displacement of a particle varies according to the relation x = 4(cosπt + sinπt). The amplitude of the particle is

(1)   8           

(2)  – 4

(3)   4          

(4)   42

Subtopic:  Simple Harmonic Motion |
 76%
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A S.H.M. is represented by x=52sin 2πt+cos 2πt. The amplitude of the S.H.M. is

(1)   10 cm         

(2)  20 cm

(3)   52 cm     

(4)  50 cm

Subtopic:  Simple Harmonic Motion |
 76%
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 Amplitude of a wave is represented by

A=ca+b-c

Then resonance will occur when

(1)     b=-c/2              

(2)         b = 0 and a c

(3)     b=-a/2              

(4)         None of these

Subtopic:  Damped Oscillations (OLD NCERT) |
 72%
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