In a spring pendulum, in place of mass, a liquid is used. If liquid leaks out continuously, then the time period of the spring pendulum:
1. | Decreases continuously |
2. | Increases continuously |
3. | First increases and then decreases |
4. | First decreases and then increases |
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. \(\dfrac{T_0}{\sqrt{2}}\)
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. | \(\frac{T}{6}\) |
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. |
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. |
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. \(T'=2T\)
1. | \(T_1<T_2\) | 2. | \(T_1>T_2\) |
3. | \(T_1=T_2\) |
4. | \(T_1= 2T_2\) |