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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
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.
More Questions from Class 11 Physics Oscillations Additional Exercises
- Q14.20: An air chamber of volume $V$ has a neck area of cross section $a$ into which a ball of mass $m$ just fits and can move up and down without any friction (Fig.14.27). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.27].
- Q14.21(a): You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant $k$ and
- Q14.21(b): You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (b) the damping constant $b$ for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
- Q14.23: A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant $\alpha$ is defined by the relation $J = –\alpha \theta$, where $J$ is the restoring couple and $\theta$ the angle of twist).
- Q14.24(a): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm
- Q14.24(b): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (b) 3 cm
- Q14.24(c): A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (c) 0 cm.
- Q14.25: A mass attached to a spring is free to oscillate, with angular velocity $\omega$, in a horizontal plane without friction or damping. It is pulled to a distance $x_0$ and pushed towards the centre with a velocity $v_0$ at time $t = 0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega$, $x_0$ and $v_0$. [Hint : Start with the equation $x = a cos (\omega t+\theta)$ and note that the initial velocity is negative.]
CBSE Solutions for Class 11 Physics Oscillations
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