Question

A harmonic wave moving in the positive x direction. Given the amplitude A=0.1 m, period T=0.5 s, wavelength λ=10 m. The original point and the source of the wave coincide. The displacement of the wave source is positive maximum at t=0. (1) Find the wave function. (2) What is the displacement of the point x= λ /4 at t1=T/4? (3) What is the vibration speed of the point x= λ /4 at t2=T/2?

Answer 1

Given that,

Amplitude A =0.1 m

Time period = 0.5 s

(I). We need to write the wave function

Using formula of wave function

$y=A\sin(kx-\omega t)$

Put the value into the formula

$y=0.1\sin(kx)$

(II). We need to calculate the displacement of the point $x=\dfrac{\lambda}{4}$ at $t=\dfrac{T}{4}$

Using equation of wave motion

$y=A\sin2\pi(\dfrac{t}{T}-\dfrac{x}{\lambda})$

Put the value in the equation

$y=0.1\sin2\pi(\dfrac{T}{4T}-\dfrac{\lambda}{4\lambda})$

$y=0.1\sin 2\pi(\dfrac{1}{4}-\dfrac{1}{4})$

$y=0\ cm$

(II). We need to calculate the vibration speed of the point x= λ /4 at t₁=T/2

Using equation of wave motion

$y=A\sin\dfrac{2\pi}{\lambda}(vt-x)$

Put the value in the equation

$y=A\sin\dfrac{2\pi}{\lambda}(v\dfrac{T}{2}-\dfrac{\lambda}{4})$

Put the value in the equation

$0=0.1\sin\dfrac{2\pi}{10}(v\times\dfrac{0.5}{2}-\dfrac{10}{4})$

$0=0.1\times0.0109(\dfrac{v}{4}-2.5)$

$0.1\times0.0109\times\dfrac{v}{4}-0.1\times0.0109\times2.5=0$

$0.0002725v-0.002725=0$

$v=\dfrac{0.002725}{0.0002725}$

$v=10\ m/s$

Hence, (I). The wave function is $y=0.1\sin(kx)$

(II). The displacement of the point is zero.

(III). The vibration speed of the point is 10 m/s.