Q. No.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
1. \(10 \) cm
2. \(15\) cm
3. \(20\) cm
4. \(25\) cm
A tuning fork has a frequency of \(200\) Hz. If the velocity of sound in air is \(330\) m/s, then how far the sound has traversed while the tuning fork completes \(20\) vibrations?
| 1. | \(11~\text{m}\) | 2. | \(22~\text{m}\) |
| 3. | \(33~\text{m}\) | 4. | \(44~\text{m}\) |
Which of the following is not possible for sound waves in air?
1. beats
2. interference
3. diffraction
4. polarization
A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is \(1.7~\text{km/s}\)? The operating frequency of the scanner is \(4.2~\text{MHz}\).
| 1. | \(3.0 \times10^{-4}~\text{m}\) | 2. | \(4.0 \times10^{-4}~\text{m}\) |
| 3. | \(3.5 \times10^{-4}~\text{m}\) | 4. | \(2.0 \times10^{-4}~\text{m}\) |
\(y(x,t)=0.06\sin\Big(\dfrac{2\pi}{3}x\Big)\cos(120\pi t)\)
where \(x\) and \(y\) are in meter and \(t\) in second. The length of the string is \(1.5~\text{m}\) and its mass is \(3\times10^{-2}~\text{kg}.\) The tension in the string is:
1. \(540~\text{N}\)
2. \(648~\text{N}\)
3. \(200~\text{N}\)
4. \(425~\text{N}\)
The wave described by \(y=0.25\sin (10\pi x-2\pi t)\), where \(x \) and \(y\) are in metre and \(t\) in second, is a wave travelling along the:
| 1. | –ve x-direction with frequency \(1\) Hz |
| 2. | +ve x-direction with frequency \(\pi\) Hz and wavelength \(\lambda=0.2\) m |
| 3. | +ve x-direction with frequency \(1\) Hz and wavelength \(\lambda=0.2\) m |
| 4. | –ve x-direction with amplitude \(0.25\) m and wavelength \(\lambda=0.2\) m |
A transverse wave is represented by \(y=A\mathrm{sin}(\omega t-kx).\)
At what value of the wavelength is the wave velocity equal to the maximum particle velocity?
1. \(\pi A/2\)
2. \(\pi A\)
3. \(2\pi A\)
4. \(A\)
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. | \(\dfrac{1}{\sqrt{\nu}} =\dfrac{1}{\sqrt{\nu_1}} +\dfrac{1}{\sqrt{\nu_2}}+\dfrac{1}{\sqrt{\nu_3}}\) |
| 1. | \( 155~\text{Hz} \) | 2. | \( 205~\text{Hz} \) |
| 3. | \( 10.5~\text{Hz} \) | 4. | \( 105~\text{Hz} \) |
| 1. | \(L\) | 2. | \(2L\) |
| 3. | \(\dfrac{L}{2}\) | 4. | \(4L\) |
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