A sinusoidal wave representing transverse oscillations on astring has the wave f
ID: 1739453 • Letter: A
Question
A sinusoidal wave representing transverse oscillations on astring has the wave function y(x,t) =(0.40m) cos[(10.0 rad/s)t - (6.00 rad/m)x] The linear mass density of the string is 0.015 kg/m. a) What are the phase velocity of the wave, the tension in thestring and the period of oscillations? b) Find the transverse velocity and acceleration of the wavewhen x = 3.60m and t = 2.00s. c) If the ends of the string are fixed, a second sinusoidalwave whose wave function is given by y(x,t)= (0.40m) cos[(10.0 rad/s)t + (6.00 rad/m)x] travels on the string. Find the superposition of thesewaves. d) Sketch the resulting interference pattern and give a cleardescription of it. A sinusoidal wave representing transverse oscillations on astring has the wave function y(x,t) =(0.40m) cos[(10.0 rad/s)t - (6.00 rad/m)x] The linear mass density of the string is 0.015 kg/m. a) What are the phase velocity of the wave, the tension in thestring and the period of oscillations? b) Find the transverse velocity and acceleration of the wavewhen x = 3.60m and t = 2.00s. c) If the ends of the string are fixed, a second sinusoidalwave whose wave function is given by y(x,t)= (0.40m) cos[(10.0 rad/s)t + (6.00 rad/m)x] travels on the string. Find the superposition of thesewaves. d) Sketch the resulting interference pattern and give a cleardescription of it.Explanation / Answer
y(x,t) = (0.40m) cos[(10.0 rad/s)t - (6.00 rad/m)x] = (0.40m) cos{(6.00 rad/m) [x - (10.0 rad/s)/(6.00 rad/m) t]} compare it with y(x, t) = A cos[2(x - ut)/] a) u = (10.0 rad/s)/(6.00 rad/m) = 1.67 m/s b) y(x,t) = (0.40m) cos[(10.0 rad/s)t - (6.00 rad/m)x] v(x,t) = dy(x, t)/dt = -(0.40m)*(10.0 rad/s) sin[(10.0 rad/s)t -(6.00 rad/m)x] so v(3.60, 2.00) = 4.00 m/s a(x,t) = dv(x, t)/dt = -(0.40m)*(10.0 rad/s)2 cos[(10.0rad/s)t - (6.00 rad/m)x] so a(3.60, 2.00) = 1.17 m/s2 c) y(x, t) = (0.40m) cos[(10.0 rad/s)t - (6.00 rad/m)x]+ (0.40m) cos[(10.0 rad/s)t + (6.00 rad/m)x] = (0.40 m) * 2 * cos[(10.0 rad/s)t] * cos[(6.00 rad/m)x] = (0.80 m) cos[(10.0 rad/s)t] cos[(6.00 rad/m)x]