Two equations of S.H.M. are y1=asin(ωt-α) and y2=bcos(ωt-α). The phase difference between the two is:
1. \(0^\circ\)
2. \(\alpha^\circ\)
3. \(90^\circ\)
4. \(180^\circ\)

Subtopic:  Simple Harmonic Motion |
 86%
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If a particle in SHM has a time period of \(0.1\) s and an amplitude of \(6\) cm, then its maximum velocity will be:
1. \(120 \pi\) cm/s 

2. \(0.6 \pi\) cm/s 

3. \(\pi\) cm/s

4. \(6\) cm/s

Subtopic:  Simple Harmonic Motion |
 90%
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If the potential energy U (in J) of a body executing SHM is given by U = 20 + 10 (sin2100πt), then the minimum potential energy of the body will be:

1. Zero 2. 30 J
3. 20 J 4. 40 J
Subtopic:  Energy of SHM |
 72%
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The kinetic energy (K) of a simple harmonic oscillator varies with displacement (x) as shown. The period of the oscillation will be: (mass of oscillator is 1 kg)

                     

1. π2 sec 2.  12 sec
3.  π sec 4. 1 sec
Subtopic:  Energy of SHM |
 75%
From NCERT
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The equation of an SHM is given as y=3sinωt + 4cosωt where y is in centimeters. The amplitude of the SHM will be?

1. 3 cm 2. 3.5 cm
3. 4 cm 4. 5 cm
Subtopic:  Linear SHM |
 90%
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The time periods for the figures (a) and (b) are T1 and T2 respectively. If all surfaces shown below are smooth, then the ratio T1T2 will be:
   

1.  1: 3

2.  1: 1

3.  2: 1

4.  3: 2

Subtopic:  Spring mass system |
 81%
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A particle is attached to a vertical spring and pulled down a distance of 0.01 m below its mean position and released. If its initial acceleration is 0.16 m/s2, then its time period in seconds will be:

1.  π

2.  π2

3.  π4

4.  2π

Subtopic:  Spring mass system |
 88%
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A particle is executing linear simple harmonic motion with an amplitude \(a\) and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be:
1. \(\frac{a\omega}{4}\)
2. \(\frac{a\omega}{2\pi}\)
3. \(\frac{2a\omega}{\pi}\)
4. \(\frac{a\omega}{\sqrt{3}\pi}\)

Subtopic:  Linear SHM |
 55%
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Two simple harmonic motions, y1 = a sin ωt and y2 = 2a sinωt + 2π3  are superimposed on a particle of mass m. The maximum kinetic energy of the particle will be:

1.  122a2

2.  52a24

3.  322a2

4.  Zero

Subtopic:  Energy of SHM |
 58%
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All the surfaces are smooth and springs are ideal. If a block of mass \(m\) is given the velocity \(v_0\) in the right direction, then the time period of the block shown in the figure will be:

                       
1. \(\frac{12l}{v_0}\)
2. \(\frac{2l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
3. \(\frac{4l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
4. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)

Subtopic:  Spring mass system |
 53%
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