Question

A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is $220m{s}^{-1}$ and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.

Answer 1

Frequency of the tunning fork,

f = 660 Hz

Wave speed, v = 220 m/s

$\Rightarrow \lambda =\frac{V}{f}=\frac{220}{660}=\frac{1}{3}m$

No. of loop s = 3

(a) So, $L.f=\left(\frac{3}{2}\right)v$

$\Rightarrow L.660=\left(\frac{3}{2}\right)\times 220$

$\Rightarrow L=\frac{1}{2}m=50cm$

(b) The equation of resultant stationary wave is given by,

$y=2Acos\left(\frac{2\pi x}{\lambda }\right)sin\left(\frac{2\pi vt}{\lambda }\right)$

$\Rightarrow y=\left(0.5\right)cmcos\left(\frac{2\pi x}{\frac{1}{3}m}\right)sin\left(\frac{2\pi \times 220\times t}{\frac{1}{3}}\right)$

$\Rightarrow y=\left(0.5cm\right)$

$cos\left(6\pi x{m}^{-1}\right)sin\left(1320\pi s^{-1}t\right)$

$\Rightarrow y=\left(0.5cm\right)$

$cos\left(0.06\pi c{m}^{-1}\right)sin\left(1320\pi s^{-1}t\right)$