Class 11 Physics - Oscillations (Important Q&A)
Oscillations Marks · 2M 3M
Click on “📋 Click for Solution” button to reveal the answer.
📘 2 Mark Q&A
3+ Q&ASolution:
The time period of a simple pendulum is given by
\( T = 2\pi \sqrt{\dfrac{l}{g}} \).
Inside a satellite orbiting the Earth, the effective acceleration due to gravity is zero because the satellite and all objects inside it are in a state of free fall.
Thus, \( g = 0 \).
Substituting in the formula, we get
\( T = 2\pi \sqrt{\dfrac{l}{0}} \), which is not defined (or tends to infinity).
Since there is no restoring force acting on the bob, it will not oscillate.
Therefore, a simple pendulum experiment cannot be performed inside a satellite.
Solution:
Damped vibration is the vibration in which the amplitude gradually decreases with time due to the presence of resistive forces such as friction or air resistance.
In such motion, a part of the mechanical energy is continuously lost as heat or other forms of energy.
As a result, the oscillations slowly die out and finally come to rest.
Solution:
In forced vibration, the amplitude of vibration depends on the frequency of the external periodic force.
As the frequency of the applied force increases, the amplitude gradually increases.
When the applied frequency becomes equal to the natural frequency of the system, the amplitude becomes maximum. This condition is called resonance.
If the frequency is increased further beyond the natural frequency, the amplitude again decreases.
Thus, the graph between amplitude (y-axis) and frequency (x-axis) is a resonance curve, which rises to a sharp maximum at the natural frequency and then falls.
Special Note: In exam, it is not necessary to write the above statements. Hurray!
Graph:
📗 3 Mark Q&A
5+ Q&ASolution:
Given displacement, \( x = A \sin \omega t + B \cos \omega t \).
Differentiate with respect to time, we get
\( \dfrac{dx}{dt} = A \omega \cos \omega t - B \omega \sin \omega t \).
Again differentiating, we get
\( \dfrac{d^2x}{dt^2} = -A \omega^2 \sin \omega t - B \omega^2 \cos \omega t \).
\( \Rightarrow \dfrac{d^2x}{dt^2} = -\omega^2 ( A \sin \omega t + B \cos \omega t ) \).
But \( A \sin \omega t + B \cos \omega t = x \).
So, \( \dfrac{d^2x}{dt^2} = -\omega^2 x \).
This is the standard differential equation of simple harmonic motion,
\( a = -\omega^2 x \).
Since acceleration is directly proportional to displacement and directed towards the mean position, the motion is simple harmonic.
Hence proved.
Solution:
Total energy of a particle executing SHM is
\( E = \dfrac{1}{2} k A^2 \)
Potential energy at displacement \(x\) is
\( U = \dfrac{1}{2} k x^2 \)
Kinetic energy at displacement \(x\) is
\( K = E - U = \dfrac{1}{2} k A^2 - \dfrac{1}{2} k x^2 \)
Given \( K = U \)
So,
\( \dfrac{1}{2} k x^2 = \dfrac{1}{2} k A^2 - \dfrac{1}{2} k x^2 \)
\( \Rightarrow k x^2 = \dfrac{1}{2} k A^2 \)
\( \Rightarrow x^2 = \dfrac{A^2}{2} \)
\( \therefore x = \pm \dfrac{A}{\sqrt{2}} \)
Answer:Therefore, the displacement from the mean position is \( \dfrac{A}{\sqrt{2}} \).
Solution:
In simple harmonic motion, displacement of a particle is
\( x = A \sin \omega t \)
Differentiating twice with respect to time,
\( \dfrac{d^2x}{dt^2} = -\omega^2 A \sin \omega t \)
But \( A \sin \omega t = x \)
So, \( \dfrac{d^2x}{dt^2} = -\omega^2 x \)
Acceleration \( a = -\omega^2 x \)
Therefore,
\( \dfrac{a}{x} = -\omega^2 \).
Since \( \omega^2 \) is constant, the ratio of acceleration to displacement remains constant.
Hence proved.
Solution:
Let the displacement in SHM be \( x = A \sin \omega t \)
As velocity is obtained by differentiating displacement with respect to (w.r.t.) time,
So, \( v = \dfrac{dx}{dt} = A\omega \cos \omega t \)
\( \therefore v = A\omega \sin \left( \omega t + \dfrac{\pi}{2} \right) \)
As acceleration is obtained by differentiating velocity w.r.t. time,
So, \( a = \dfrac{d^2x}{dt^2} = -A\omega^2 \sin \omega t \)
\( \therefore a = A\omega^2 \sin \left( \omega t + \pi \right) \)
Phase of velocity \( = \omega t + \dfrac{\pi}{2} \)
Phase of acceleration \( = \omega t + \pi \)
Therefore, phase difference between velocity and acceleration is
\( \left( \omega t + \pi \right) - \left( \omega t + \dfrac{\pi}{2} \right) = \dfrac{\pi}{2} \)
Hence, the phase difference between velocity and acceleration is \( \dfrac{\pi}{2} \) (i.e., \(90^\circ\)).
Solution:
| Forced Vibration | Resonance |
|---|---|
| When a body vibrates under the influence of an external periodic force, it is called forced vibration. | Resonance is a special case of forced vibration. |
| The body vibrates with the frequency of the external force. | It occurs when the frequency of the external force becomes equal to the natural frequency of the body. |
| The amplitude depends on the frequency of the external force and damping. | At resonance, the amplitude of vibration becomes maximum. |
| It occurs for any frequency of the external force. | It occurs only at one particular frequency (natural frequency). |
Hi Please, do not Spam in Comments.