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

\[\text{ Wave length,} \lambda = \frac{v}{f} = \frac{220}{660} = \frac{1}{3}  m\] 

(a) No. of loops, n = 3 
∴  \[L = \frac{n}{2}\lambda\]

\[\Rightarrow L = \frac{3}{2} \times \frac{1}{3}\] 

\[ \Rightarrow L = \frac{1}{2}  m = 50  \text{ cm }\]
(b) Equation of resultant stationary wave can be given by:

\[y = 2A\cos\left( \frac{2\pi x}{\lambda} \right)\sin\left( \frac{2\pi vL}{\lambda} \right)\] 

\[ \Rightarrow y = 0 . 5  \cos\left( \frac{2\pi x}{\frac{1}{3}} \right)\sin\left( \frac{2\pi \times 220 \times t}{\frac{1}{3}} \right)\] 

\[ \Rightarrow y = 0 . 5  cm  \cos\left( 6\pi x  m^{- 1} \right)  \sin\left( 1320\pi t  s^{- 1} \right)\]