Question

A sinusoidal wave with wavelength 0.300 m travels along a string. The maximum transverse speed of a point on the string is 3.00 m/s and the maximum transverse acceleration is 8.50×104m/s2. 1) What is the propagation speed v of the wave?

Answer 1

Propagation speed is given by,

v = f*

here, = wavelength = 0.300 m

f = frequency = w/(2*pi)

given, maximum transverse speed of a point on the string(V_max) = A*w = 3.00 m/s

maximum transverse acceleration(a_max) = A*w^2 = 8.50*10^4 m/s^2

then, a_max/V_max = w = (8.50*10^4)/3.00

w = 2.833*10^4 rad/sec.

then, f = (2.833*10^4)/(2*pi)

So,  propagation speed of the wave,

v = (2.833*10^4)/(2*pi)*0.300

v = 1.35*10^3 m/s

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