A sinusoidal transverse wave is traveling along a string toward decreasing x . T

Question

A sinusoidal transverse wave is traveling along a string toward decreasing x. The figure below shows a plot of the vertical (y) displacement of the string as a function of position along the string (x) at time t = 0. The x axis is marked in increments of 5 cm and the y axis is marked in increments of 1 cm. The string tension is 4 N, and its linear density is 24 g/m.

(a) Find the amplitude.
5cm

(b) Find the wavelength.
.40m

(c) Find the wave speed.
12.91m/s

(d) Find the period of the wave.
30.98ms

(e) Find the maximum speed of a particle in the string.
10.166m/s

(f) Find the maximum size of the transverse acceleration of a particle on the string, and indicate when it occurs:

2067.34 m/s2

(f) Complete the equation describing the traveling wave, in whichx and y are in meters and t is inseconds.y(x, t) = .05m sin( 15.7 m-1 x +203.34 s-1 t + .927rad)

vy(x,t) = (10.166m/s) cos ( _?_ m-1 x (+/-?) _?_ s-1 t + _?_ rad )

ay(x,t) = (_?_m/s2) sin ( _?_ m-1 x (+/-?) _?_ s-1 t + _?_ rad )

Explanation / Answer

a) amplitude = A = 5 cm

b)

wavelength = 4*5 = 20 cm

c)

v = sqrt(T/u)

v = sqrt(4/0.024) = 13 m/s

(d)

T = wavelength/v = 0.2/13 = 15.4 ms

(e)

Vmax = A*w = A*2*pi/T

Vmax = 0.05*2*pi/0.0154 = 20.4 m/s

(f)

a = w^2*A = 4*pi2*A/T2 = 0.05*4*pi2/0.01542 = 8323.2 m/s2

(g)

Vy = 10.66m/s cos (15.7m-1 + 203.34 s-1 + 0.927 rad )

amax = w*Vy = 203.34*10.66 = 2167.6 m/s2

ay = -2167.6 m/s2 sin(15.7 m-1 + 203.34s-1 + 0.927 rad )

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