Equation of a simple harmonic motion is given by \(x= a\sin \omega t\). For which value of \(x\), kinetic energy is equal to the potential energy?
1. \(x = \pm a\)
2. \(x = \pm \frac{a}{2}\)
3. \(x = \pm \frac{a}{\sqrt{2}}\)
4. \(x = \pm \frac{\sqrt{3}a}{2}\)

Subtopic:  Energy of SHM |
 82%
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A spring pendulum is placed on a rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) If the the angular velocity of the table becomes \(2\omega_{0},\) then the new time period of the pendulum will be:
1. 
\(2T_{0}\)
2. \(T_0\sqrt{2}\)
3. the same
4. \(\dfrac{T_0}{\sqrt{2}}\)

Subtopic:  Angular SHM |
 71%
From NCERT
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The displacement \( x\) of a particle varies with time \(t\) as \(x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)\)The time taken by the particle to reach from \(x = \frac{A}{2} \) to \(x = -\frac{A}{2} \) will be:

1. \(\frac{T}{2}\) 2. \(\frac{T}{3}\)
3. \(\frac{T}{12}\) 4. \(\frac{T}{6}\)

Subtopic:  Phasor Diagram |
From NCERT
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Force on a particle \(F\) varies with time \(t\) as shown in the given graph. The displacement \(x\) vs time \(t\) graph corresponding to the force-time graph will be:
          

1. 2.
3. 4.
Subtopic:  Linear SHM |
 66%
From NCERT
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A particle executes SHM with a frequency of \(20~\text{Hz}\). The frequency with which its potential energy oscillates is:
1. \(5~\text{Hz}\)
2. \(20~\text{Hz}\)
3. \(10~\text{Hz}\)
4. \(40~\text{Hz}\)
Subtopic:  Energy of SHM |
 74%
From NCERT
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The graph of potential energy \((U)\) versus displacement \((x)\) is shown. Which of the following describes the oscillation about the mean position, \(x = 0\text{?}\)

1. 2.
3. 4.
Subtopic:  Energy of SHM |
 88%
From NCERT
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A spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)

Subtopic:  Combination of Springs |
 68%
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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:  Linear SHM |
 71%
From NCERT
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Two simple pendulums of length \(1~\text{m}\) and \(16~\text{m}\) are in the same phase at the mean position at any instant. If \(T\) is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:
1. \(\frac{3T}{2}\)
2. \(\frac{3T}{4}\)
3. \(\frac{2T}{3}\)
4. \(\frac{4T}{3}\)
Subtopic:  Angular SHM |
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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. \(2~\text{s}\)
Subtopic:  Linear SHM |
 75%
From NCERT
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