Q. No.The equation of an SHM is given as \(y = 3\sin\omega t+ 4\cos \omega t\) where \(y\) is in centimeters. The amplitude of the SHM will be?
| 1. | \(3~\text{cm}\) | 2. | \(3.5~\text{cm}\) |
| 3. | \(4~\text{cm}\) | 4. | \(5~\text{cm}\) |
Subtopic: Â Linear SHM |
 90%
From NCERT
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A particle executing simple harmonic motion of amplitude \(5~\text{cm}\) has a maximum speed of \(31.4~\text{cm/s}.\) The frequency of its oscillation will be:
1. \(1~\text{Hz}\)
2. \(3~\text{Hz}\)
3. \(2~\text{Hz}\)
4. \(4~\text{Hz}\)
Subtopic: Â Linear SHM |
 86%
From NCERT
AIPMT - 2005
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Two simple harmonic motions of angular frequency \(100~\text{rad s}^{-1}\) and \(1000~\text{rad s}^{-1}\) have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. \(1:10\)
2. \(1:10^{2}\)
3. \(1:10^{3}\)
4. \(1:10^{4}\)
Subtopic: Â Linear SHM |
 86%
From NCERT
NEET - 2008
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If a particle is executing SHM, with an amplitude \(A\), the distance moved and the displacement of the body in a time equal to its time period are, respectively:
| 1. | \(2A,A\) | 2. | \(4A,0\) |
| 3. | \(A,A\) | 4. | \(0,2A\) |
Subtopic: Â Linear SHM |
 83%
From NCERT
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A body performs simple harmonic motion about \(x=0\) with an amplitude a and a time period \(T\). The speed of the body at \(x= \frac{a}{2}\) will be:
1. \(\frac{\pi a\sqrt{3}}{2T}\)
2. \(\frac{\pi a}{T}\)
3. \(\frac{3\pi^2 a}{T}\)
4. \(\frac{\pi a\sqrt{3}}{T}\)
1. \(\frac{\pi a\sqrt{3}}{2T}\)
2. \(\frac{\pi a}{T}\)
3. \(\frac{3\pi^2 a}{T}\)
4. \(\frac{\pi a\sqrt{3}}{T}\)
Subtopic: Â Linear SHM |
 77%
From NCERT
NEET - 2009
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The amplitude of a simple harmonic oscillator is \(A\) and speed at the mean position is \(v_0\). The speed of the oscillator at the position \(x={A \over \sqrt{3}}\) will be:
| 1. | \(2v_0 \over \sqrt{3}\) | 2. | \(\sqrt{2}v_0 \over 3\) |
| 3. | \({2 \over 3}v_0\) | 4. | \(\sqrt{\frac{2}{3}}v_0\) |
Subtopic: Â Linear SHM |
 77%
From NCERT
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A particle executes SHM with a time period of \(4~\text{s}\). The time taken by the particle to go directly from its mean position to half of its amplitude will be:
1. \(\frac{1}{3}~\text{s}\)
2. \(1~\text{s}\)
3. \(\frac{1}{2}~\text{s}\)
4. \(2~\text{s}\)
1. \(\frac{1}{3}~\text{s}\)
2. \(1~\text{s}\)
3. \(\frac{1}{2}~\text{s}\)
4. \(2~\text{s}\)
Subtopic: Â Linear SHM |
 75%
From NCERT
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A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)
2. \( \frac{T}{6} \)
3. \(\frac{T}{3} \)
4. \( \frac{T}{12}\)
Subtopic: Â Linear SHM |
 70%
From NCERT
AIPMT - 2008
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A particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is \(T_1\) and to go from \(A/2\) to \(A\) is \(T_2\) then:
| 1. | \(T_1<T_2\) | 2. | \(T_1>T_2\) |
| 3. | \(T_1=T_2\) |
4. | \(T_1= 2T_2\) |
Subtopic: Â Linear SHM |
 71%
From NCERT
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The amplitude and the time period in an SHM are \(0.5\) cm and \(0.4\) sec respectively. If the initial phase is \(\frac{\pi}{2}\) radian, then the equation of SHM will be:
1. \(y = 0.5\sin(5\pi t)\)
2. \(y = 0.5\sin(4\pi t)\)
3. \(y = 0.5\sin(2.5\pi t)\)
4. \(y = 0.5\cos(5\pi t)\)
1. \(y = 0.5\sin(5\pi t)\)
2. \(y = 0.5\sin(4\pi t)\)
3. \(y = 0.5\sin(2.5\pi t)\)
4. \(y = 0.5\cos(5\pi t)\)
Subtopic: Â Linear SHM |
 69%
From NCERT
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