Two pendulums suspended from the same point have lengths of \(2\) m and \(0.5\) m. If they are displaced slightly and released, then they will be in the same phase when the small pendulum has completed:
1. \(2\) oscillations
2. \(4\) oscillations
3. \(3\) oscillations
4. \(5\) oscillations
1. | \(0.01~\text{Hz}\) | 2. | \(0.02~\text{Hz}\) |
3. | \(0.03~\text{Hz}\) | 4. | \(0.04~\text{Hz}\) |
1. | \(2n\) | 2. | \(n/2\) |
3. | \(n\) | 4. | none of the above |
A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:
1. \(2 \pi \sqrt{\frac{\left(\right. M + m \left.\right) l}{Mg}}\)
2. \(2 \pi \sqrt{\frac{ml}{Mg}}\)
3. \(2 \pi \sqrt{L / g}\)
4. \(2 \pi \sqrt{\frac{Ml}{\left(\right. m + M \left.\right) g}}\)
The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite
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 total energy of the particle performing SHM depends on:
1. \(k,\) \(a,\) \(m\)
2. \(k,\) \(a\)
3. \(k,\) \(a\), \(x \)
4. \(k,\) \(x \)
When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:
1.
2.
3.
4.
The time period of a mass suspended from a spring is \(T\). If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
1. \(\frac{T}{4}\)
2. \(T\)
3. \(\frac{T}{2}\)
4. \(2T\)