A sinusoidal wave is traveling on a string with speed 20. cm/s. The displacement

Question

A sinusoidal wave is traveling on a string with speed 20. cm/s. The displacement of the particles of the string at x = 25 cm is found to vary with time according to the equation y = (5.0 cm) sin[10.0 - (8.0 s-1)t]. The linear density of the string is 7.0 g/cm.
(a) What is the frequency of the wave?
____s-1

(b) What is the wavelength of the wave?
_____ cm

(c) Give the general equation giving the transverse displacement of the particles of the string as a function of position and time.
y(x,t) = ( ____ cm) sin[( ____cm-1)x - ( ____s-1)t]

(d) What is the tension in the string?
_____ N

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Explanation / Answer

y = (5.0 cm) sin[10.0 - (8.0 s-1)t]
the general expression for the wave is: y(x,t) = ym sin(kx -t)

at x=25cm

y(x,t) = ym sin(k*25 -t)

compare with the expression: y = (5.0 cm) sin[10.0 - (8.0 s-1)t]

=8 rad/sec

f = /(2) = 1.27388 Hz

Since 25k = 10, then k=10/25 =0.4

Wave length = (2)/k = 6.28/0.4 =15.7cm

The amplitute ym = 5 cm

By substituting the values of k and into the general expression y(x,t) = (5.0 cm) sin[10.0 - (8.0 s-1)t]

since v=/k = (/)

the tension is =2 /k2

[(8^2)*7g/cm]/(0.4)^2 = 0.02800 N

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