Q. No.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. \(2T\)
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. | \(\sqrt{T} \) | 2. | \(T \) |
| 3. | \({T}^{1 / 3} \) | 4. | \(\sqrt{2} {T}\) |
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. | \(\dfrac{T_0}{\sqrt{2}}\) |
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. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)
The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite
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. |  |
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. \(2\pi\sqrt{\frac{L}{g\tan\theta}}\)
| 1. | \(T_2 ~\text{is infinity} \) | 2. | \(T_2>T_1 \) |
| 3. | \(T_2<T_1 \) | 4. | \(T_2=T_1\) |
1. \(\frac{3T}{2}\)
2. \(\frac{3T}{4}\)
3. \(\frac{2T}{3}\)
4. \(\frac{4T}{3}\)
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