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Q14.22:
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Solution :

Step 1: Establishing the Equations of Motion

For a particle undergoing Simple Harmonic Motion (SHM) along a straight line, the displacement $x(t)$ as a function of time is given by:

$x(t) = A \sin(\omega t + \phi)$

[Per the definition of SHM, where $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is the phase constant].

The instantaneous velocity $v(t)$ is the first derivative of displacement with respect to time:

$v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$

Step 2: Defining Kinetic and Potential Energy

The instantaneous Kinetic Energy ($K$) is defined as:

$K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi)$

[Using the relation $k = m\omega^2$, where $k$ is the force constant, we can write $K = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)$].

The instantaneous Potential Energy ($U$) is defined as:

$U = \frac{1}{2}kx^2 = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$

Step 3: Calculating Average Energy Over One Period ($T$)

The average value of a function $f(t)$ over a period $T$ is given by $\langle f \rangle = \frac{1}{T} \int_{0}^{T} f(t) dt$.

For Kinetic Energy:

$\langle K \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \cos^2(\omega t + \phi) dt$

Using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$\langle K \rangle = \frac{kA^2}{2T} \int_{0}^{T} \frac{1 + \cos(2(\omega t + \phi))}{2} dt$

Since the integral of $\cos(2\omega t + 2\phi)$ over a full period $T = \frac{2\pi}{\omega}$ is zero:

$\langle K \rangle = \frac{kA^2}{4T} [t]_{0}^{T} = \frac{kA^2}{4T} \cdot T = \frac{1}{4}kA^2$

Step 4: Calculation for Potential Energy

Similarly, for Potential Energy:

$\langle U \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \sin^2(\omega t + \phi) dt$

Using the identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$:

$\langle U \rangle = \frac{kA^2}{2T} \int_{0}^{T} \frac{1 - \cos(2(\omega t + \phi))}{2} dt$

$\langle U \rangle = \frac{kA^2}{4T} [t]_{0}^{T} = \frac{1}{4}kA^2$

Visual Representation of Energy Distribution

Time (t) Energy K(t) U(t)

Conclusion

Comparing the derived averages:

Energy Type Average Value
Kinetic Energy ($\langle K \rangle$) $\frac{1}{4}kA^2$
Potential Energy ($\langle U \rangle$) $\frac{1}{4}kA^2$

Final Solution: Since $\langle K \rangle = \frac{1}{4}kA^2$ and $\langle U \rangle = \frac{1}{4}kA^2$, it is proven that $\langle K \rangle = \langle U \rangle$ for a particle in linear SHM.


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