(1) A wave on a string under tension T=4.1 N whose total mass is 4 gms and lengt

Question

(1) A wave on a string under tension T=4.1 N whose total mass is 4 gms and length
is 3 m is described by
y(x,t) = 15 sin (kx – 4p t)
where x and y are in centimeters and t is in seconds.
(a) What is the angular frequency of the transverse oscillations of a point on the
string?
(b) What is the speed of the wave moving along the string and in what direction is it
moving?
(c) What is the transverse speed for a point on the string at x= 6.0 cm when t is 0.25
s?
(d) What is the maximum transverse speed of any point on the string?
(e) What is the maximum transverse acceleration for any point on the string?

Explanation / Answer

Given that Tension in the string, T = 4.1 N The mass of the string, m = 4 g = 0.004 kg The length of the sting is, l = 3 m Linear mass density, = m/l = 0.0013 kg/m And the given equation is y(x,t) = 15 sin (kx – 4 t) a) By comparing the above equation with the following y = A sin (kx - t) Then we get The angular frequency, = 4 = 12.56 rad/s b) The speed of the wave, v = sqrt (T/) = 56.16 m/s c) We have, k = /v = 12.56/56.16 = 0.224 So the given equation can be written as y(x,t) = 15 sin (0.224x – 4 t) By differentiating the above we get v = dy(x,t)/dt = (0.224x – 4 t)15 cos (0.224x – 4 t) For the given variables, x = 6 cm = 0.06 m and time t = 0.25 s Then v = (0.224*0.06 – 4 * 0.25)15 cos (0.224*0.06 – 4*0.25) v = 46.8 m/s d) The maximum transverse speed, Vmax = A = 15 * 12.56 = 188.4 m/s e) The maximum transverse acceleration, amax = A2 = 2366.3 m/s^2 y(x,t) = 15 sin (0.224x – 4 t) By differentiating the above we get v = dy(x,t)/dt = (0.224x – 4 t)15 cos (0.224x – 4 t) For the given variables, x = 6 cm = 0.06 m and time t = 0.25 s Then v = (0.224*0.06 – 4 * 0.25)15 cos (0.224*0.06 – 4*0.25) v = 46.8 m/s d) The maximum transverse speed, Vmax = A = 15 * 12.56 = 188.4 m/s e) The maximum transverse acceleration, amax = A2 = 2366.3 m/s^2 v = dy(x,t)/dt = (0.224x – 4 t)15 cos (0.224x – 4 t) For the given variables, x = 6 cm = 0.06 m and time t = 0.25 s Then v = (0.224*0.06 – 4 * 0.25)15 cos (0.224*0.06 – 4*0.25) v = 46.8 m/s d) The maximum transverse speed, Vmax = A = 15 * 12.56 = 188.4 m/s e) The maximum transverse acceleration, amax = A2 = 2366.3 m/s^2 v = (0.224*0.06 – 4 * 0.25)15 cos (0.224*0.06 – 4*0.25) v = 46.8 m/s d) The maximum transverse speed, Vmax = A = 15 * 12.56 = 188.4 m/s e) The maximum transverse acceleration, amax = A2 = 2366.3 m/s^2 v = 46.8 m/s d) The maximum transverse speed, Vmax = A = 15 * 12.56 = 188.4 m/s e) The maximum transverse acceleration, amax = A2 = 2366.3 m/s^2    

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