Question

A string of length L fixed at both ends vibrates in its fundamental mode at a frequency $\nu$ and a maximum amplitude A. (a) Find the wavelength and the wave number k. (b). Take the origin at one end of the string and the X-axis along the string. Take the Y-axis along the direction of the displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-direction. Write the equation describing the standing wave.

Answer 1

Given:
Length of the string = L
Velocity of wave is given as:
$V=\sqrt{\frac{T}{m}}$
In fundamental mode, frequency = ν
$\Rightarrow \nu =\frac{1}{2L}\sqrt{\frac{T}{m}}$
(a) $\mathrm{Wavelength},\lambda =\frac{\text{V}\mathrm{elocity}}{\text{F}\mathrm{requency}}$
$$\begin{gathered} \Rightarrow \lambda =\frac{\sqrt{{\displaystyle \frac{T}{m}}}}{{\displaystyle \frac{1}{2L}}\sqrt{{\displaystyle \frac{T}{m}}}}=2L \\ \mathrm{Wave}\mathrm{number},K=\frac{2\mathrm{\pi }}{\lambda }=\frac{2\mathrm{\pi }}{2L}=\frac{\mathrm{\pi }}{L} \end{gathered}$$

(b) Equation of the stationary wave:
$$\begin{gathered} y=A\cos \left(\frac{2\pi x}{\lambda }\right)\sin \left(\frac{2\pi Vt}{\lambda }\right) \\ =A\cos \left(\frac{2\pi x}{2L}\right)\sin \left(\frac{2\pi Vt}{2L}\right)\left[\because \nu =\left(\frac{V}{2L}\right)\right] \\ =A\cos \left(\frac{\pi x}{L}\right)\sin \left(2\pi \nu t\right) \end{gathered}$$