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 circular motion of a particle with constant speed is:
1. | Periodic and simple harmonic | 2. | Simple harmonic but not periodic |
3. | Neither periodic nor simple harmonic | 4. | Periodic but not simple harmonic |
1. | \(2n\) | 2. | \(n/2\) |
3. | \(n\) | 4. | none of the above |
1. | The value of \(a\) is zero whatever may be the value of \(v\). |
2. | When \(v\) is zero, \(a\) is zero. |
3. | When \(v\) is maximum, \(a\) is zero. |
4. | When \(v\) is maximum, \(a\) is maximum. |
Two springs of spring constants \(k_1\) and \(k_2\) are joined in series. The effective spring constant of the combination is given by:
1. \(\frac{k_1+k_2}{2}\)
2. \(k_1+k_2\)
3. \(\frac{k_1k_2}{k_1+k_2}\)
4. \(\sqrt{k_1k_2}{}\)
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
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 total energy of the particle performing SHM depends on:
1. \(k,\) \(a,\) \(m\)
2. \(k,\) \(a\)
3. \(k,\) \(a\), \(x \)
4. \(k,\) \(x \)