Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Two identical waves, each with amplitude \(X_{0}\) and intensity \(I\), interfere constructively. What are the amplitude and intensity of the resultant wave?
Correct Answer: \( \boxed{\mathrm{D}} \)
For constructive interference, the two waves are in phase, so their amplitudes add directly:
\( \displaystyle A_{\mathrm{result}}=A_{1}+A_{2}. \)
Since the waves are identical,
\( \displaystyle A_{1}=A_{2}=X_{0}, \)
giving
\( \displaystyle A_{\mathrm{result}}=X_{0}+X_{0}=2X_{0}. \)
Wave intensity is proportional to the square of the amplitude:
\( \displaystyle I\propto A^{2}. \)
Therefore,
\( \displaystyle I_{\mathrm{result}}=\left(\frac{2X_{0}}{X_{0}}\right)^{2}I=4I. \)
Hence, the resultant wave has an amplitude of \( \displaystyle 2X_{0} \) and an intensity of \( \displaystyle 4I. \)
Therefore, the correct answer is (D).
Two wave pulses approach each other, as shown in the figure above. The pulse traveling to the right is half the height and twice the width of the pulse traveling to the left. Which of the following best represents the resulting shape when the peaks of the two pulses are at the same location?
Correct Answer: \( \boxed{\mathrm{D}} \)
The shape of the resulting pulse is determined by the principle of superposition, which states that the displacement of the medium at any point is the algebraic sum of the displacements due to the individual pulses.
At the instant the peaks coincide:
• The upward pulse has amplitude \(+1\).
• The downward pulse has amplitude \(-0.5\) and is twice as wide.
The overlapping region therefore has a net displacement of
\( \displaystyle 1-0.5=0.5. \)
Since the negative pulse is wider, its outer portions remain below the equilibrium position where there is no overlap, while the central region has a reduced positive peak.
This produces a smaller positive pulse with negative “wings” on either side, matching the shape shown in option (D).
Therefore, the correct answer is (D).
A \(340\,\mathrm{Hz}\) tuning fork sets an air column vibrating in fundamental resonance, as shown above. A hollow tube is inserted into a column of water, and the height of the tube is adjusted until strong resonance is heard. At approximately what length of the air column will this happen?
Correct Answer: \( \boxed{\mathrm{D}} \)
The tube is closed at one end by the water surface and open at the other end, so the fundamental resonance occurs when the air-column length is one-quarter of the wavelength:
\( L=\frac{\lambda}{4} \)
First, calculate the wavelength using
\( \lambda=\frac{v}{f} \)
where \(v=340\,\mathrm{m/s}\) and \(f=340\,\mathrm{Hz}\).
\( \lambda=\frac{340}{340}=1.0\,\mathrm{m} \)
Therefore,
\( L=\frac{1.0}{4}=0.25\,\mathrm{m}=25\,\mathrm{cm} \)
Thus, the air column resonates most strongly when its length is approximately \(25\,\mathrm{cm}\), so the correct answer is (D).
