Class 11 Physics - Waves (Important Q&A)
Waves Marks · 2M 3M 4M
Click on “📋 Click for Solution” button to reveal the answer.
📗 3 Mark Q&A
6+ Q&ASolution:
Given wave equation \( y = 20 \sin (0.2\pi x - 50\pi t) \).
Comparing with standard form \( y = A \sin (kx - \omega t) \),
Wave number \( k = 0.2\pi \).
Angular frequency \( \omega = 50\pi \).
We know \( k = \dfrac{2\pi}{\lambda} \).
So, \( 0.2\pi = \dfrac{2\pi}{\lambda} \).
\( \Rightarrow \lambda = \dfrac{2\pi}{0.2\pi} \).
\( \therefore \lambda = 10 \text{ cm} \).
Wave velocity \( v = \dfrac{\omega}{k} \).
\( \Rightarrow v = \dfrac{50\pi}{0.2\pi} \).
\( \therefore v = 250 \text{ cm/s} \).
Answer: The required wavelength \( \lambda = 10 \text{ cm} \) and velocity \( v = 250 \text{ cm/s} \).
Solution:
Given wave equation \( y = 4 \sin 20\pi \left( t - \dfrac{x}{100} \right) \).
\( \Rightarrow y = 4 \sin \left( 20\pi t - \dfrac{20\pi}{100} x \right) \)
\( \therefore y = 4 \sin \left( 20\pi t - 0.2\pi x \right) \)
Comparing with standard form \( y = A \sin (\omega t - kx) \), we get
Angular frequency \( \omega = 20\pi \)
We know \( \omega = 2\pi f \).
So, \( 20\pi = 2\pi f \).
\( \Rightarrow f = \dfrac{20\pi}{2\pi} \).
\( \therefore f = 10 \text{ Hz} \).
Answer: The required frequency of the wave is \( 10 \text{ Hz} \).
Solution:
Given wavelength \( \lambda = 2 \, \text{m} \).
Distance between two particles \( x = 1 \, \text{m} \)
Phase difference is given by \( \Delta \phi = \dfrac{2\pi}{\lambda} \times x \)
Substituting the vakue of \(x \), we get \( \Delta \phi = \dfrac{2\pi}{2} \times 1 \)
\( \therefore \Delta \phi = \pi \text{ rad} \)
Answer: The phase difference between the two particles is \( \pi \) rad (or \(180^\circ\)).
Solution:
When a sound wave passes from one medium to another, the frequency of the wave remains unchanged.
This is because the frequency is determined by the source of the sound and does not depend on the medium.
However, the speed and wavelength change according to the properties of the new medium.
Since \( v = f\lambda \), if velocity \(v\) changes and frequency \(f\) remains constant, then wavelength \( \lambda \) must change.
Therefore, the physical quantity that remains unchanged is the frequency of the wave.
Solution:
The laws of transverse vibration of a stretched string are as follows:
(1) Law of Length: The frequency \(f\) of vibration is inversely proportional to the length \(L\) of the string, when tension and mass per unit length are constant.
\( f \propto \dfrac{1}{L} \).
(2) Law of Tension: The frequency \(f\) is directly proportional to the square root of the tension \(T\) in the string, when length and mass per unit length are constant.
\( f \propto \sqrt{T} \).
(3) Law of Mass: The frequency \(f\) is inversely proportional to the square root of the mass per unit length \( \mu \), when length and tension are constant.
\( f \propto \dfrac{1}{\sqrt{\mu}} \).
On combining all three laws, we get
\( f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}} \).
Solution:
Given \( y_1 = a_1 \sin \omega t \) and \( y_2 = a_2 \cos \omega t \)
Resultant displacement \( y = y_1 + y_2 \).
\( y = a_1 \sin \omega t + a_2 \cos \omega t \)
Assume \( a_1 = A \cos \phi \) and \( a_2 = A \sin \phi \)
Substituting in the expression for \(y\),
\( y = A \cos \phi \sin \omega t + A \sin \phi \cos \omega t \)
\( \Rightarrow y = A ( \sin \omega t \cos \phi + \cos \omega t \sin \phi ) \)
As we all know that \( \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \),
So, \( y = A \sin (\omega t + \phi) \)
Thus the resultant wave has amplitude \(A\).
Since \( a_1 = A \cos \phi \) and \( a_2 = A \sin \phi \),
\( \Rightarrow a_1^2 + a_2^2 = A^2 (\cos^2 \phi + \sin^2 \phi) \).
\( \Rightarrow a_1^2 + a_2^2 = A^2 \).
\( \therefore A = \sqrt{a_1^2 + a_2^2} \).
Answer: The amplitude of the resultant wave is \( \sqrt{a_1^2 + a_2^2} \).
Solution:
For an open organ pipe, all harmonics are present and
\( f_n = n f_1 \), where \(n = 1, 2, 3, \dots \)
Given third harmonic,
\( f_3 = 3 f_1 = 900 \, \text{Hz} \).
So, \( f_1 = \dfrac{900}{3} \).
\( \Rightarrow f_1 = 300 \, \text{Hz} \).
Therefore, the frequency of fifth harmonic,
\( f_5 = 5 f_1 \).
\( \Rightarrow f_5 = 5 \times 300 \).
\( \therefore f_5 = 1500 \, \text{Hz} \).
Answer: The frequency of the fifth harmonic is \(1500 \, \text{Hz}\).
Hi Please, do not Spam in Comments.