Question

A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.

Answer 1

Explanation:

Given, Length of String (L) = 2.0 m

Wave speed of the string in its first overtone (v) = 200 $ms^-^1$

Amplitude (A) = 0.5 cm  

(a) Wavelength and frequency of the string when it is vibrating in its First Overtone (n = 2)-

L = $\frac {n\lambda}{2}$

$\lambda$ = L = 2m

f = $\frac {\nu}{\lambda} = \frac {200}{2}$ = 100 Hz

(b) The stationary wave equation is given by -

y = $2A cos \frac {2\pi x}{\lambda} sin \frac {2\pi vt}{\lambda}$

  = $0.5 cos \frac {2\pi x}{2} sin \frac {2\pi 200t}{2}$

  = $(0.5cm) cos [(\pi m^-^1)x] sin[(200\pi s^-^1)t]$