Question

20. Energy in waves. As a sine-shaped wave moves along a stretched string, each coil os the spring will execute the simple harmonic motion. Suppose that a spring with linear density µ= 0.40 kg/m has a tension T=1.6N in it and that a sine-shaped wave of amplitude 10cm and wavelength of 1.5 is moving along the sprig. Consider a 1cm segment of the spring.

a. What is the mass of the 1cm segment of the spring?

b. What is the speed of the wave along the spring?

c. What is the period of the simple harmonic motion executed by this segment of the spring?

d. Knowing the mass of the segment and the period of its simple harmonic motion, find the effective spring constant of the coil spring for transverse displacements away from its equilibrium.

e. What is the total energy of the simple harmonic oscillator of the segment?

f. There are 150 1cm segments within the 1.5-meter wavelength. What will be the total harmonic oscillation energy in one wavelength of the wave?

g. At what rate in joules/seconds (or with what power in watts) will this energy be transferred past a given point on the spring?

Answer 1

a) Given mass per unit length = µ = 0.40 kg/m
For 1metre the mass is 0.40kg,
For 1cm segment mass is 0.40/100 kg = 0.004kg

b) Speed of the wave v is given by, v = where T is the tension in the spring, and µ is mass/length
v = = = 2 m/s

c) The amplitude of SHM is 10cm and the wavelength is 1.5 m. The velocity of the wave is 2m/s as calculated above.
Frequency of wave is given by formula f = v/ , where is the wavelength
So f = 2 / 1.5 = 4/3 Hz
The time period is the inverse of frequency T = 1/f = 1/(4/3) = 3/4 = 0.75 s

d)
T =   , where m is mass, and k is the spring constant

0.75 =
1.178 =
1.387 = 0.004 /k
Spring constant , k = 0.004 / 1.387 = 0.0028 N/m


e) Amplitude of SHM = 10cm = 0.1 m
The energy of SHM of segment of 1cm= maximum potential energy = 1/2 k A² = 1/2 * 0.0028 * (0.1)² = 0.000014 Joules

f) Total Harmonic Oscillation energy of 150 segments or 1 wavelength = 150 * Energy of 1 segment calculated above
= 150 * 0.000014 J = 0.0021 Joules

g) Power transferred = Energy associated with one wavelength / Time period = 0.0021 / 0.75 = 0.0028 J/s or Watts