The transverse displacement of a string fixed at both ends is given by $y = 0.06\sin{\left( \frac{2\pi x}{3} \right)\cos{(120\pi t)}}\ y$ and $x$ are in metres and $t$ in seconds. The wavelength and frequency of the two superposing waves are

Question

Dr. Abubacker · Accepted Answer

(d) Waves travelling to the right can be given by
$y_{1} = A\sin{(\omega t - kx)}$ …(i)
When getting reflected from the fixed end of the string, there is an additional phase difference of $\pi.$ The reflected wave is
$y_{2} = A\sin{(\omega t + kx + \pi)}$
$\Rightarrow y_{2} = - A\sin(\omega t + kx)$ …(ii)
Superposing, (i) + (ii) is the same as $y = \sin{C - \sin D}$
$y = 2A\cos\frac{C + D}{2}\sin\frac{C - D}{2}$
$y = 2A\cos{\omega t}\sin{kx}$
The stationary wave is given as
$y = 0.06\sin\frac{2\pi x}{3}\cos{(120\pi t)}$
Here $k = \frac{2\pi}{\lambda} = \frac{2\pi}{3}\ $ and $\omega = 120\pi$
$	herefore\lambda = 3m,\ v = \frac{120\pi}{2\pi} = 60Hz$

Q. The transverse displacement of a string fixed at both ends is given by $y = 0.06\sin{\left( \frac{2\pi x}{3} \right)\cos{(120\pi t)}}\ y$ and $x$ are in metres and $t$ in seconds. The wavelength and frequency of the two superposing waves are

Explanation

Reviewed by Dr. Abubacker

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