A spring pendulum is on the rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) Now the angular velocity of the table becomes \(2\omega_{0},\) then the new time period will be:
1. \(2T_{0}\)
2. \(T_0\sqrt{2}\)
3. remains the same
4. \(\frac{T_0}{\sqrt{2}}\)
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 curve between the potential energy \((U)\) and displacement \((x)\) is shown. Which of the oscillation is about the mean position, \(x = 0?\)
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. | \(T_1= 2T_2\) |
Two simple pendulums of length 1 m and 16 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.
2.
3.
4.