There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,F=-Kx where x is the displacement. The total energy of body depends upon -

(1)   K, x         

(2)  K, a

(3)   K, a, x    

(4)  K, a, v

Subtopic:  Energy of SHM |
 71%
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The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

(1)    18E       

(2)    14E

(3)    12E       

(4)    23E

Subtopic:  Energy of SHM |
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A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

(1)  P.E. is maximum when x = 0

(2)  K.E. is maximum when x = 0

(3)  T.E. is zero when x = 0

(4)  K.E. is maximum when x is maximum

Subtopic:  Energy of SHM |
 89%
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­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration g4 , then the period of the pendulum will be

(1) T

(2) T4

(3) 2T5

(4) 2T5

Subtopic:  Angular SHM |
 85%
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The total energy of a particle, executing simple harmonic motion is:
1. \(\propto x\)
2. \(\propto x^2\)
3. Independent of \(x\)
4. \(\propto x^{\frac{1}{2}}\)
Subtopic:  Energy of SHM |
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A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by T=2πlg',  where g'   is equal to

(1) g                                                       

(2) g-a

(3) g+a

(4) g2+a2

Subtopic:  Angular SHM |
 88%
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If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day?

1. 3927 sec

2. 3727 sec

3. 3427 sec

4. 864 sec

Subtopic:  Simple Harmonic Motion |
 68%
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In a simple pendulum, the period of oscillation T is related to length of the pendulum l as

(1) lT=constant

(2) l2T=constant

(3) lT2=constant

(4) l2T2=constant

Subtopic:  Angular SHM |
 83%
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The equation of motion of a particle is \({d^2y \over dt^2}+Ky=0 \) where \(K\) is a positive constant. The time period of the motion is given by: 
1. \(2 \pi \over K\) 2. \(2 \pi K\)
3. \(2 \pi \over \sqrt{K}\) 4. \(2 \pi \sqrt{K}\)
Subtopic:  Simple Harmonic Motion |
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The kinetic energy of a particle executing SHM is \(16~\text{J}\) when it is in its mean position. If the amplitude of oscillations is \(25~\text{cm}\) and the mass of the particle is \(5.12~\text{kg}\), the time period of its oscillation will be:
1. \(\frac{\pi}{5}~\text{s}\)
2. \(2\pi ~\text{s}\)
3. \(20\pi ~\text{s}\)
4. \(5\pi ~\text{s}\)
Subtopic:  Energy of SHM |
 71%
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