Q. No.The phase difference between two waves, represented by
\(y_1= 10^{-6}\sin \left\{100t+\left(\frac{x}{50}\right) +0.5\right\}~\text{m}\)
\(y_2= 10^{-6}\cos \left\{100t+\left(\frac{x}{50}\right) \right\}~\text{m}\)
where \(x\) is expressed in metres and \(t\) is expressed in seconds, is approximate:
1. \(2.07\) radians
2. \(0.5\) radians
3. \(1.5\) radians
4. \(1.07\) radians
\(y_1= 10^{-6}\sin \left\{100t+\left(\frac{x}{50}\right) +0.5\right\}~\text{m}\)
\(y_2= 10^{-6}\cos \left\{100t+\left(\frac{x}{50}\right) \right\}~\text{m}\)
where \(x\) is expressed in metres and \(t\) is expressed in seconds, is approximate:
1. \(2.07\) radians
2. \(0.5\) radians
3. \(1.5\) radians
4. \(1.07\) radians
| 1. | \({y}=0.2 \sin \left[2 \pi\left(6{t}+\frac{x}{60}\right)\right]\) |
| 2. | \({y}=0.2 \sin \left[ \pi\left(6{t}+\frac{x}{60}\right)\right]\) |
| 3. | \({y}=0.2 \sin \left[2 \pi\left(6{t}-\frac{x}{60}\right)\right]\) |
| 4. | \(y=0.2 \sin \left[ \pi\left(6{t}-\frac{x}{60}\right)\right]\) |
1. \(3:2\)
2. \(2:3\)
3. \(9:4\)
4. \(4:9\)
1. \(\frac{1}{n}= \frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \(n= n_1\times n_2\times n_3\)
3. \(n= n_1+ n_2+ n_3\)
4. \(n= \frac{n_1+ n_2+ n_3}{3}\)
1. \(1\%\)
2. \(2\%\)
3. \(3\%\)
4. \(4\%\)
\(y=(0.02~\text{m})\sin\left[{{2}\pi \left({\frac{t}{{0.04}~(\text{s})}-\frac{x}{{0.50}~(\text{m})}}\right)}\right]\). The tension in the string will be:
| 1. | \(4.0~\text{N}\) | 2. | \(12.5~\text{N}\) |
| 3. | \(0.5~\text{N}\) | 4. | \(6.25~\text{N}\) |
1. \(324.16~\text{m/s}\)
2. \(320~\text{m/s}\)
3. \(345~\text{m/s}\)
4. \(346.8~\text{m/s}\)
A bat emits an ultrasonic sound of frequency \(1000\) kHz in the air. If the sound meets a water surface, what is the wavelength of the reflected sound? (The speed of sound in air is \(340\) m/sec and in water is \(1486\) m/sec)
1. \(3.4 \times 10^{-4}~\text{m}\)
2. \(1 . 49 \times 10^{- 3} ~ \text{m}\)
3. \(2 . 34 \times 10^{- 2} ~\text{m}\)
4. \(1 . 73 \times10^{- 3} ~\text{m}\)
A steel wire has a length of \(12.0\) m and a mass of \(2.10\) kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air, at \(20^{\circ}\text{C}\) (which is \(343\) m/s)?
1. \(4.3\times10^3\) N
2. \(3.2\times10^4\) N
3. \(2.06\times10^4\) N
4. \(1.2\times10^4\) N
1. \(\frac{v(t_1-t_2)}{2}\)
2. \(\frac{v(t_1t_2)}{2(t_1+t_2)}\)
3. \(v(t_1+t_2)\)
4. \(\frac{v(t_1+t_2)}{2}\)
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