A string is fixed at both ends and set to vibrate in five loops. If the wavelength is \(8\) cm then the length of the string is:
1. \(10 \) cm
2. \(15\) cm
3. \(20\) cm
4. \(25\) cm
If the speed of sound in air is \(v,\) then the minimum possible length of the closed-end organ pipe which resonates to frequency \(f\) will be:
| 1. | \(\dfrac{v}{2f}\) | 2. | \(\dfrac{v}{4f}\) |
| 3. | \(\dfrac{v}{3f}\) | 4. | \(\dfrac{v}{f}\) |
| 1. | \(30^{\circ}\) | 2. | \(45^{\circ}\) |
| 3. | \(60^{\circ}\) | 4. | \(90^{\circ}\) |
When a string is divided into three segments of lengths \(l_1\), \(l_2\) and \(l_3\), the fundamental frequencies of these three segments are \(\nu_1\), \(\nu_2\) and \(\nu_3\) respectively. The original fundamental frequency (\(\nu\)) of the string is:
| 1. | \(\sqrt{\nu} = \sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}\) |
| 2. | \(\nu = \nu_1+\nu_2+\nu_3\) |
| 3. | \(\dfrac{1}{\nu} =\dfrac{1}{\nu_1} +\dfrac{1}{\nu_2}+\dfrac{1}{\nu_3}\) |
| 4. | \(\dfrac{1}{\sqrt{\nu}} =\dfrac{1}{\sqrt{\nu_1}} +\dfrac{1}{\sqrt{\nu_2}}+\dfrac{1}{\sqrt{\nu_3}}\) |
A pipe is \(15\) cm long and is open at both ends. Which harmonic mode of the pipe will resonate with a \(2.2\) kHz sound source?
(Given: the velocity of sound in air \(=330\) m/s)
| 1. | fundamental | 2. | second harmonic |
| 3. | third harmonic | 4. | fourth harmonic |
| 1. | \( 155~\text{Hz} \) | 2. | \( 205~\text{Hz} \) |
| 3. | \( 10.5~\text{Hz} \) | 4. | \( 105~\text{Hz} \) |
| (a) | Every particle has a fixed amplitude which is different from the amplitude of its nearest particle. |
| (b) | All the particles cross their mean position at the same time. |
| (c) | All the particles are oscillating with same amplitude. |
| (d) | There is no net transfer of energy across any plane. |
| (e) | There are some particles which are always at rest. |