# Homework Help: Energy in standing waves

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1. May 10, 2016

### Titan97

1. The problem statement, all variables and given/known data
The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment, the displacement of the wire is given by $y=A\sin\left(\frac{\pi x}{L}\right)\sin(\omega t)$ and its energy is $E_1$. In another experiment, the displacement of wire is given by $y=A\sin\left(\frac{2\pi x}{L}\right)\sin(2\omega t)$ and its energy is $E_2$. If $E_2=kE_1$, find $k$ (k is a positive integer).

2. Relevant equations
None

3. The attempt at a solution
$$y=A\sin\left(\frac{\pi x}{L}\right)\sin(\omega t)=\frac{A}{2}\left[\cos\left(\frac{\pi x}{L}-\omega t\right)-\cos\left(\frac{\pi x}{L}+\omega t\right)\right]$$

The given wave is formed by two waves travelling in opposite direction with amplitudes $\frac{A}{2}$. At $t=0$, the two waves completely cancel out each other. Hence the energy becomes zero. Isnt $E_1$ time dependent?

2. May 10, 2016

### Orodruin

Staff Emeritus
The waves only cancel each other in the displacement. This means the potential is zero at that moment. They do not cancel in the velocity. In fact, the string has its maximal velocity at that time.

I suggest finding an expression for the kinetic energy of the string and compute it at t=0 when the potential is zero.

No, the energies are not time dependent.

3. May 10, 2016

### Titan97

Using $\text{d}K=\frac{1}{2}(\mu \text{d}x) v_y^2$
$$K=\frac{1}{4}\mu\omega^2A^2L\cos^2\omega t$$

at $t=0$
$$E_1=\frac{1}{4}\mu\omega^2A^2L$$
$$E_2=\frac{1}{4}\cdot 2\mu\cdot 4\omega^2A^2\frac{L}{2}=4E_1$$

4. May 10, 2016

### Titan97

@Orodruin how can I compute potential energy as a function of time?

5. May 11, 2016

### Orodruin

Staff Emeritus
You have to compare the length of the string to the rest length. The potential energy is the tension multiplied by this change in length.

To get energy conservation with the waves you will need to use the first order approximation, but that was also used for deriving the wave equation so it is consistent.

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