# 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 - Physics

Sum

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.

#### Solution

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:

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

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

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

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 43 | Page 326