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 220 m 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

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

$\text{Wavelength,}\mathrm{\lambda }=\frac{v}{\mathrm{f}}=\frac{220}{660}=\frac{1}{3}\mathrm{m}$
(a) No. of loops, n = 3
∴ $L=\frac{n}{2}\lambda$
$$\begin{gathered} \Rightarrow L=\frac{3}{2}\times \frac{1}{3} \\ \Rightarrow L=\frac{1}{2}\mathrm{m}=50\mathrm{cm} \end{gathered}$$

(b) Equation of resultant stationary wave can be given by:
$$\begin{gathered} y=2A\cos \left(\frac{2\pi x}{\lambda }\right)\sin \left(\frac{2\pi vL}{\lambda }\right) \\ \Rightarrow y=0.5\cos \left(\frac{2\pi x}{{\displaystyle \frac{1}{3}}}\right)\sin \left(\frac{2\pi \times 220\times t}{{\displaystyle \frac{1}{3}}}\right) \\ \Rightarrow y=0.5\mathrm{cm}\cos \left(6\pi x{m}^{-1}\right)\sin \left(1320\pi t{\mathrm{s}}^{-1}\right) \end{gathered}$$