Q. No.A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector \(\vec a\) is correctly shown in:
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A simple pendulum of mass \(m\) swings about point \(B\) between extreme positions \(A\) and \(C\). Net force acting on the bob at these three points is correctly shown by:
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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. | \(\sqrt{2} {T}\) |
| 1. | \(T_2 ~\text{is infinity} \) | 2. | \(T_2>T_1 \) |
| 3. | \(T_2<T_1 \) | 4. | \(T_2=T_1\) |
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. \(\frac{3T}{2}\)
2. \(\frac{3T}{4}\)
3. \(\frac{2T}{3}\)
4. \(\frac{4T}{3}\)
A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 , where y is in meters and t is in seconds. If g = 10 , then the time period of the pendulum will be:
| 1. | 4 s | 2. | 6 s |
| 3. | 2 s | 4. | 12 s |
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
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