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. \(k,\) \(x \)
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. T/4
2. T
3. T/2
4. 2T
A particle of mass \(m\) oscillates with simple harmonic motion between points \(x_1\) and \(x_2\), the equilibrium position being \(O\). Its potential energy is plotted. It will be as given below in the graph:
1. | 2. | ||
3. | 4. |
The potential energy of a simple harmonic oscillator, when the particle is halfway to its endpoint, will be:
1. \(\frac{2E}{3}\)
2. \(\frac{E}{8}\)
3. \(\frac{E}{4}\)
4. \(\frac{E}{2}\)
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. |