If the length of a pendulum is made \(9\) times and mass of the bob is made \(4\) times, then the value of time period will become:
1. \(3T\)
2. \(\dfrac{3}{2}T\)
3. \(4T\)
4. \(2T\)

Subtopic:  Angular SHM |
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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. \(\frac{L^2}{T^2}= \text{constant}\)
Subtopic:  Angular SHM |
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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 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 |
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Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then:
1. \(l_A = \frac{l_B}{4}\)
2. \(l_A= 4l_B\)
3. \(l_A= 2l_B~\&~M_A=2M_B\)
4. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)

Subtopic:  Angular SHM |
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The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

Subtopic:  Angular SHM |
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A simple pendulum is pushed slightly from its equilibrium towards the left and then set free to execute the simple harmonic motion. Select the correct graph between its velocity (\(v\)) and displacement (\(x \)).

1.   2.
3.   4.
Subtopic:  Angular SHM |
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The period of oscillation of a simple pendulum of length \(L\) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination \(\theta\), is given by:
1. \(2\pi\sqrt{\frac{L}{g\cos\theta}}\)
2. \(2\pi\sqrt{\frac{L}{g\sin\theta}}\)
3. \(2\pi\sqrt{\frac{L}{g}}\)
4. \(2\pi\sqrt{\frac{L}{g\tan\theta}}\)

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 |
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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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