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. \(\sqrt{\mathrm{T}} \) 2. \(T \)
3. \(\mathrm{T}^{1 / 3} \) 4. \(\sqrt{2} \mathrm{~T}\)

Subtopic:  Angular SHM |
 82%
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A simple pendulum hanging from the ceiling of a stationary lift has a time period T1. When the lift moves downward with constant velocity, then the time period becomes T2. It can be concluded that: 

1. \(T_2 ~\text{is infinity} \) 2. \(\mathrm{T}_2>\mathrm{T}_1 \)
3. \(\mathrm{T}_2<\mathrm{T}_1 \) 4. \(T_2=T_1\)
Subtopic:  Angular SHM |
 60%
From NCERT
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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. 3/2T

3. 4T

4. 2T

Subtopic:  Angular SHM |
 83%
From NCERT
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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 |
 86%
From NCERT
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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 |
 83%
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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=2πReg

(2) T=2π2Reg

(3) T=2πRe2g

(4) T=2 seconds

Subtopic:  Simple Harmonic Motion |
 69%
From NCERT
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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%
From NCERT
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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) |
From 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%
From NCERT
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