Q.A progressive wave travelling along a stretched string is described by $y=0.02\sin(4x-200t)$ (SI units). Find

The maximum PARTICLE velocity in a travelling wave (not the wave speed itself) is A x omega, read off directly from the wave equation's amplitude and angular frequency.

Given: y = 5 sin(100 pi t - 2 pi x), with x, y in metres and t in seconds.

Comparing with the standard progressive wave equation y = A sin(omega t - kx):

Amplitude, A = 5 m

Angular frequency, omega = 100 pi rad/s

(wave number k = 2 pi rad/m, though not needed for this particular question)

The transverse velocity of a particle of the medium is:

v_particle = dy/dt = A omega cos(omega t - kx)

This is maximum when the cosine term equals 1: …

For a wave y = A sin(omega t - kx), the particle (transverse) velocity is dy/dt = A omega cos(...), whose maximum magnitude is A omega.

Given wave equation: y = 5 sin(100 pi t - 2 pi x), so comparing with the standard form y = A sin(omega t - k x):

Amplitude A = 5 m.

Angular frequency omega = 100 pi rad/s.

Particle velocity: v_p = dy/dt = A omega cos(100 pi t - 2 pi x).

Maximum particle velocity = A omega = 5 x 100 pi = 500 pi m/s.

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