Average value of KE and PE over entire time period is

Options

A. `0,1/2momega^2A^2`
B. `1/2momega^2A^2,0`
C. `1/2momega^2A^2,1/2momega^2A^2`
D. `1/4momega^2A^2,1/4momega^2A^2`

Answer

To solve the problem of finding the average value of kinetic energy (KE) and potential energy (PE) over an entire time period for a wave, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Wave Equation**:
The wave can be described by the equation \( y = A \sin(\omega t - kx) \). Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

2. **Kinetic Energy Expression**:
The kinetic energy of a particle in SHM (Simple Harmonic Motion) can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the particle. The velocity can be derived from the displacement:
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t - kx) \]
Therefore, substituting \( v \) into the kinetic energy expression:
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t - kx))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t - kx) \]

3. **Average Kinetic Energy Calculation**:
To find the average kinetic energy over one complete cycle (from \( 0 \) to \( \frac{2\pi}{\omega} \)), we use:
\[ \langle KE \rangle = \frac{1}{T} \int_0^T KE \, dt \]
where \( T = \frac{2\pi}{\omega} \). Thus, we have:
\[ \langle KE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t - kx) \, dt \]
The average of \( \cos^2 \) over one complete cycle is \( \frac{1}{2} \):
\[ \langle KE \rangle = \frac{1}{2} m A^2 \omega^2 \cdot \frac{1}{2} = \frac{1}{4} m A^2 \omega^2 \]

4. **Potential Energy Expression**:
The potential energy in SHM can be expressed as:
\[ PE = \frac{1}{2} k x^2 \]
For a wave, we can relate \( k \) to the amplitude and angular frequency. The average potential energy over one complete cycle is also:
\[ \langle PE \rangle = \frac{1}{4} m A^2 \omega^2 \]
because, in SHM, the average potential energy is equal to the average kinetic energy.

5. **Final Result**:
Therefore, the average values of kinetic energy and potential energy over the entire time period are:
\[ \langle KE \rangle = \langle PE \rangle = \frac{1}{4} m A^2 \omega^2 \]