16.4 Energy and power of a wave  (Page 3/6)

The time-averaged power of a sinusoidal mechanical wave, which is the average rate of energy transfer associated with a wave as it passes a point, can be found by taking the total energy associated with the wave divided by the time it takes to transfer the energy. If the velocity of the sinusoidal wave is constant, the time for one wavelength to pass by a point is equal to the period of the wave, which is also constant. For a sinusoidal mechanical wave, the time-averaged power is therefore the energy associated with a wavelength divided by the period of the wave. The wavelength of the wave divided by the period is equal to the velocity of the wave,

Note that this equation for the time-averaged power of a sinusoidal mechanical wave shows that the power is proportional to the square of the amplitude of the wave and to the square of the angular frequency of the wave. Recall that the angular frequency is equal to $\omega =2\pi f$ , so the power of a mechanical wave is equal to the square of the amplitude and the square of the frequency of the wave.

Power supplied by a string vibrator

Strategy

Solution

1. Begin with the equation of the time-averaged power of a sinusoidal wave on a string:
   
   $P=\frac{1}{2}\mu A^2{\omega }^{2}v.$
   
   The amplitude is given, so we need to calculate the linear mass density of the string, the angular frequency of the wave on the string, and the speed of the wave on the string.
2. We need to calculate the linear density to find the wave speed:
   
   $\mu =\frac{m_s}{L_s}=\frac{0.070\text{kg}}{2.00\text{m}}=0.035\text{kg/m}.$
3. The wave speed can be found using the linear mass density and the tension of the string:
   
   $v=\sqrt{\frac{F_T}{\mu }}=\sqrt{\frac{90.00\text{N}}{0.035\text{kg/m}}}=50.71\text{m/s}.$
4. The angular frequency can be found from the frequency:
   
   $\omega =2\pi f=2\pi \left(60{\text{s}}^{\mathrm{-1}}\right)=376.80{\text{s}}^{\mathrm{-1}}.$
5. Calculate the time-averaged power:
   
   $P=\frac{1}{2}\mu A^2{\omega }^{2}v=\frac{1}{2}\left(0.035\frac{\text{kg}}{\text{m}}\right){\left(0.040\text{m}\right)}^2{\left(376.80{\text{s}}^{\mathrm{-1}}\right)}^2\left(50.71\frac{\text{m}}{\text{s}}\right)=201.59\text{W}.$

Significance