Question

A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at x=2 cm is found at its mean position and it is moving in positive y-direction at t=1 s. the amplitude of the wave, the wavelength and the angular frequency of the wave are 0.1m,pi//4m and 4pi rad//s, respectively. The speed of particle at x=2 m and t=1 s is

Answer 1

Given :

A sinusoidal wave is propagating in -ve X direction in a stretched string .

The mean position of a particle of a string is found at :

x = 2 cm , t = 1 sec and moving in +ve Y direction .

The amplitude of the wave :

= 0.1 m

Wavelength of the wave :

$\lambda = \frac{\pi}{4} m$

Angular frequency of the wave :

$\omega= 4\pi$ rad per sec

To Find :

The speed of the particle at x = 2 m and t = 1 sec  = ?

Solution :

We know that the formula for the speed of the given particle will be given as :

v  = ωA ,

where ω is angular velocity and A is amplitude of the wave.

So , according to above formula :

v= $4\pi \frac{rad}{s} \times o.1 m$

 $= 4\pi \times \frac{1}{10} \frac{m}{s^{2} } \\= 1.256 \frac{m}{s^{2} }$

So the velocity of the above given wave at x = 2 and t = 1 s  is 1.256 m per square sec