Question

A $2m$-long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is $200m{s}^{-1}$ and amplitude is $0.5cm$.

(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 $x=50cm$ has reached its maximum displacement.

Answer 1

Step 1: Given data

Length of the string, $L=2m$

The wave speed on the string, $v=200m{s}^{-1}$

Amplitude of the wave, $A=0.5cm$

$=0.5\times {10}^{-2}m\left[As1m=100cm\right]$

Vibrating in first overtone means, $n=2$

Step 2: (a) Find wavelength, $\lambda$ and frequency, $f$

Relation between length of the wire and wavelength, $L=n\lambda /2$

$$\begin{gathered} \lambda =2L/n \\ =\left(2\times 2\right)/2 \\ =2m \end{gathered}$$

Relation between frequency, wavelength and speed, $f=v/\lambda$

Frequency of the wave, $f=200m{s}^{-1}/2m$

$=100Hz$

Step 3: (b) To write the equation for the stationary wave

The stationary wave equation, $y=A\cos (2\pi x/\lambda )\times \sin (2\pi vt/\lambda )$

$$\begin{gathered} =0.5\cos (2\pi x/2)\sin \left(2\pi \right(200)t/\lambda ) \\ =\left(0.5cm\right)\cos \left[\right(\pi m^{-1}\left)x\right]\sin \left[\right(200\left)\pi s^{-1}t\right] \end{gathered}$$

Hence, wavelength, $\lambda$ and frequency, $f$ of the wave are $2m$ and $100Hz$ respectively.

Equation for the stationary wave is $\left(0.5cm\right)\cos \left[\right(\pi m^{-1}\left)x\right]\sin \left[\right(200\left)\pi s^{-1}t\right]$ .