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Q. No.A simple pendulum has a time period \(T_1\) when on the earth’s surface, and \(T_2\) when taken to a height \(R\) above the earth’s surface, where \(R\) is the radius of the earth. The value of \(\frac{T_2}{T_1}\) is:
1. \(1\)
2. \(\sqrt{2}\)
3. \(4\)
4. \(2\)
Subtopic: Simple Harmonic Motion |
51%
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The displacement of a particle along the \(x\text-\)axis is given by \(x= a\sin^2\omega t\). The motion of the particle corresponds to:
| 1. | simple harmonic motion of frequency \(\frac{\omega}{\pi}\). |
| 2. | simple harmonic motion of frequency \(\frac{3\omega}{2\pi}\). |
| 3. | non-simple harmonic motion. |
| 4. | simple harmonic motion of frequency \(\frac{\omega}{2\pi}\). |
Subtopic: Simple Harmonic Motion |
From NCERT
NEET - 2010
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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: Linear SHM |
77%
From NCERT
NEET - 2009
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Which one of the following equations of motion represents simple harmonic motion? (where \(k,k_0,k_1~\text{and}~a\) are all positive.)
1. Acceleration \(=-k_0x+k_1x^2\)
2. Acceleration \(=-k(x+a)\)
3. Acceleration \(=k(x+a)\)
4. Acceleration \(=kx\)
1. Acceleration \(=-k_0x+k_1x^2\)
2. Acceleration \(=-k(x+a)\)
3. Acceleration \(=k(x+a)\)
4. Acceleration \(=kx\)
Subtopic: Simple Harmonic Motion |
73%
From NCERT
NEET - 2009
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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: Linear SHM |
86%
From NCERT
NEET - 2008
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An SHM has an amplitude \(a\) and a time period \(T.\) The maximum velocity will be:
1. \({4a \over T}\)
2. \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)
1. \({4a \over T}\)
2. \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)
Subtopic: Simple Harmonic Motion |
90%
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A particle is executing simple harmonic motion with frequency \(f\). The frequency at which its kinetic energy changes into potential energy, will be:
1. \(\frac{f}{2}\)
2. \(f\)
3. \(2f\)
4. \(4f\)
1. \(\frac{f}{2}\)
2. \(f\)
3. \(2f\)
4. \(4f\)
Subtopic: Energy of SHM |
61%
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In a simple pendulum, the period of oscillation \(T\) is related to length of the pendulum \(L\) as:
1. \(\frac{L}{T}= \text{constant}\)
2. \(\frac{L^2}{T}= \text{constant}\)
3. \(\frac{L}{T^2}= \text{constant}\)
4. \(\frac{L^2}{T^2}= \text{constant}\)
1. \(\frac{L}{T}= \text{constant}\)
2. \(\frac{L^2}{T}= \text{constant}\)
3. \(\frac{L}{T^2}= \text{constant}\)
4. \(\frac{L^2}{T^2}= \text{constant}\)
Subtopic: Angular SHM |
84%
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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. | \(\sqrt{T} \) | 2. | \(T \) |
| 3. | \({T}^{1 / 3} \) | 4. | \(\sqrt{2} {T}\) |
Subtopic: Angular SHM |
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A simple pendulum hanging from the ceiling of a stationary lift has a time period \(T_1\). When the lift moves downward with constant velocity, then the time period becomes \(T_2\). It can be concluded that:
| 1. | \(T_2 ~\text{is infinity} \) | 2. | \(T_2>T_1 \) |
| 3. | \(T_2<T_1 \) | 4. | \(T_2=T_1\) |
Subtopic: Angular SHM |
61%
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