Consider a loop in the standing wave created by two waves (amplitude 5.00 mmand frequency 120 Hz ) traveling in opposite directions along a string with length 2.25 mand mass125gand under tension 40 N. At what rate does energy enter the loop from (a) each side and (b) both sides? (c) What is the maximum kinetic energy of the string in the loop during its oscillation?

i) Amplitude of the standing wave, $y_m=5.00mmor0.005m$

ii) Frequency of the standing wave, f =120 Hz

iii) Length of the string, L =2.25 m

iv) Mass of the string, M = 125 g or 0.125 kg

v) Tension in the string is T = 40N

The energy associated with one wavelength on a string is transmitted down the string at the propagation velocity v as a sinusoidal wave sweeps down the string.

We can find the speed and the angular speed on a string using the corresponding formulae. Inserting these in the formula for the rate at which energy is transmitted by a sinusoidal wave on a string, we get the rate at which energy enters the loop from one side. From this, we can find the rate for both sides easily. Using the relation between this and K.E, we can find the Maximum K.E of the string in the loop during the oscillation.

Formulae:

The angular frequency of a body in motion,$\omega =2\mathrm{\pi F}$ …(i)

The transverse speed of a body,$v=\sqrt{\frac{T}{\mu }}$ …(ii)

The average rate of power,$P_{avg}=\frac{1}{2}\mu v{\omega }^2{y}_m^2$ …(iii)

The rate change of kinetic energy, $\frac{dk}{dt}=\frac{1}{2}\mu v{\omega }^2{y}_m^2\left(kx-\omega t\right)$ …(iv)

Where, $\mu$is mass per unit length, $\omega$is angular speed, and k is wave number

The mass per unit length of the string is given as:

$$\begin{gathered} \mu =\frac{M}{L} \\ =\frac{0.125kg}{2.25m} \\ =0.055\frac{kg}{m} \end{gathered}$$

The speed of the string using equation (ii) is given as:

$$\begin{gathered} v=\sqrt{\frac{40N}{0.055{\displaystyle \frac{kg}{m}}}} \\ =26.96m/s \end{gathered}$$

The angular speed of the string using equation (i) is given as:

$$\begin{gathered} \omega =2\left(3.142\right)\left(120Hz\right) \\ =754.08rad/s \end{gathered}$$

The rate at which the energy is transmitted by a sinusoidal wave on a string from one side is given using equation (iii) and the given values as:

$$\begin{gathered} P_{avg}=\frac{1}{2}\left(0.055\frac{kg}{m}\right)\left(26.96m/s\right){\left(754.08rad/s\right)}^2{\left(0.005m\right)}^2 \\ =10.539W \\ \approx 10.54W \end{gathered}$$

Therefore, the rate at which energy enters the loop from one side is 10.54 W

The rate at which the energy enters the loop from both the sides is given as:

$$\begin{gathered} P=2P_{avg} \\ =2\left(10.54W\right) \\ =21.08W \end{gathered}$$

Hence, the rate at which energy enters the loop from both the sides is 21.08 W

Integrating both the sides of the equation (iv), we get the kinetic energy of the string in oscillation as:

$$\begin{gathered} K={\int }_0^T\mu v{\omega }^2{y}_m^2\left(kx-\omega t\right)dt \\ =\mu v{\omega }^2{y}_m^2{\int }_0^T\left(kx-\omega t\right)dt \\ =\frac{1}{2}\mu v{\omega }^2{y}_m^2{\int }_0^T\left(kx-\omega t\right)dt \\ =\frac{1}{2}\mu v{\omega }^2{y}_m^2\left(\frac{T}{2}\right) \\ =P_{avg}\frac{T}{2} \\ =\frac{P_{avg}}{2f} \\ =\frac{10.54W}{2\left(120Hz\right)} \\ =0.0439J \\ ~4.4\times {10}^{-2}J \end{gathered}$$ (from equation (iii))

Therefore, the maximum K.E of the string in the loop during the oscillation is $4.4\times {10}^{-2}J$