PHYS 2133
Chapter 16 Problems

Selected Solutions

  1. For transverse waves on a taut string, we know that the wave speed is given by v = (F/m)1/2, where F is the tension in the string and m is the linear mass density of the string. With the information given, we can calculate

    m = m/L = (.0600 kg)/(5 m) = .012 kg/m.

    Using this information, we find that the tension in the string is

    F = v2 m = (50 m/s)2 (.012 kg/m) = 30 N

  2. For waves travelling on a string, we know that the wave speed is given by

    v = (T/m)1/2

    For a wave speed of 20.0 m/s and a tension of 6.0 N, we find that the mass per unit length of the string is

    m = T/v2 = 6.0 N/400 m2/s2 = .015 kg/m

    In order to get a waves speed of 30.0 m/s, we need the tension to be

    T = v2m = (900 m2/s2)(.015 kg/m) = 13.5 N.

  3. To calculate how far away the mountain is, we need to know the speed of sound. Since the air is at T = 22 oC = 295 K, we know from the book that

    v = 331 m/s (1 + (T - 273 K)/273 K)1/2 = (331 m/s)(1 + (295 K - 273 K)/273 K)1/2 = 344 m/s

    Please note that the equation in the book is poorly written and should be re-written as above. Since the echo took 3.0 s to return, this means that the sound took 1.5 s to reach the mountain. At 344 m/s, this means that the mountain is

    d = (344 m/s)(1.5 s) = 516 m
     

  4. Since the frequency of the sound that is heard by the bicyclist is lower in frequency than the original sound, we know that the car must be moving away from the bicyclist, i.e. the bicyclist is behind the car, moving at a speed that is 1/3 that of the cars. Behind the car, we know that the wavelength of the sound is given by

    l' = l + vs/f

    This corresponds to a frequency that is given by

    f' = v/l' = f v/(v + vs)

    This is the frequency relative to still air. As the bicyclist moves through this wave, she hears a frequency that is increased above this still air frequency by

    f'' = f'(v + vo)/v = f (v + vo)/(v + vs)

    If we use the fact that 3vo = vs, we get

    f'' = f (v + vo)/(v + 3vo)

    Solving this for vo, we get

    f'' (v + 3vo) = f (v + vo)
    vo (3f'' - f) = v (f - f'')
    vo = v (f - f'')/(3f'' - f) = 345 m/s (25/805) = 10.7 m/s

    Since the car is moving 3 times faster, the speed of the car is 32.1 m/s.