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A transverse sinusoidal wave is moving along a string in the positive direction

ID: 1561481 • Letter: A

Question

A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 80 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.3 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 20 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± t + ), what are (c) ym, (d) k, (e) , (f) , and (g) the correct choice of sign in front of ?

Explanation / Answer

(a) The simple harmonic motion relation

= um/ym = 20/0.043= 465.11 rad/s

f = / 2 = 465.11/ 2 = 74.06Hz

(b) Using v = f , we find

= 80/74.06= 1.08 m

(c) The amplitude of the transverse displacement is

ym = 4.ecm = 0.043m

(d) The wave number is

k = 2p / = 2p / 1.08= 5.81 rad/m

(e) The angular frequency, as obtained in part (a), is = 465.11rad/s

(f) The function describing the wave can be written as

y = 0.043 sin (5.81x – 465.11t + )

where distances are in meters and time is in seconds. We adjust the phase constant f to satisfy the condition y = 0.043 at x= t= 0. Therefore, sin f= 1, for which the “simplest” root is = /2. Consequently, the answer is

y = 0.04e sin (5.81x – 465.11t + /2)

(g) The sign in front of is minus

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