Question

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of ï­1= 0.0128(kg/m). Itâs displacement function is D(x,t) = Acos(kx - ï·t). Itâs amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with ï­2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and ï·, for the original wave, as well as k and ï· the reflected wave and the transmitted wave.

Answer 1

Answer:

Explanation:

For original wave,

Given: $D(x,t) = Acos(kx - \omega t)$

A=amplitude of incident wave=0.001m

V=speed of wave

$\mu_1$=linear mass density

T=tension in string

$K=\frac{2\pi}{\lambda_1}=\frac{2\pi}{0.8}=7.854m^{-1}$

$\omega=VK=K\sqrt{\frac{T}{\mu _{1}}}=K\sqrt{\frac{8.0}{0.0128}}=196.35rad/s$

For reflected wave and transmitted wave ω remains same because frequency does not change. So

$\omega = 196.35rad/s$

For reflected wave $K=\frac{2\pi}{\lambda_1}=\frac{2\pi}{0.8}=7.854m^{-1}$

For transmitted wave $K=\frac{2\pi}{\lambda_2}\\\\\frac{V_1}{V_2}=\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mu_2}{\mu_1}}\\\\\lambda_2=0.8\sqrt{\frac{0.0128}{0.0572}}=0.4\\\\K=\frac{2\pi}{0.4}=15.708m^{-1}$