Correct answer Carries: 4.
Wrong Answer Carries: -1.
Two strings produce beats of 6 Hz. One has a frequency of 384 Hz. When the tension in the second string is decreased, the beat frequency becomes 4 Hz. What was the original frequency of the second string?
Let \( v_2 \) be the original frequency.
\( |384 - v_2| = 6 \Rightarrow v_2 = 378 \, \text{Hz or } 390 \, \text{Hz} \).
Decreasing tension decreases frequency. If \( v_2 = 390 \), new \( v_2’ < 390 \), beat = \( 390 - 384 = 6 \) or decreases, so \( v_2’ = 380 \), beat = \( 384 - 380 = 4 \), consistent.
Thus, \( v_2 = 390 \, \text{Hz} \).
Two strings produce beats of 6 Hz. One has a frequency of 300 Hz. When the tension in the second string is decreased, the beat frequency becomes 4 Hz. What was the original frequency of the second string?
\( |300 - v_2| = 6 \Rightarrow v_2 = 294 \, \text{Hz or } 306 \, \text{Hz} \).
Decreasing tension decreases frequency. If \( v_2 = 306 \), new \( v_2’ < 306 \), beat = \( 306 - 300 = 6 \) or decreases, so \( v_2’ = 296 \), beat = \( 300 - 296 = 4 \), consistent.
Thus, \( v_2 = 306 \, \text{Hz} \).
In a longitudinal wave traveling through a gas, what is the relationship between displacement nodes and pressure antinodes?
In longitudinal waves, displacement nodes (zero displacement) occur where particles are maximally compressed or rarefied, leading to maximum pressure variation (pressure antinodes).
A stationary wave is given by \( y = 0.08 \sin (\frac{\pi x}{3}) \cos (60\pi t) \), where \( x \) and \( y \) are in meters and \( t \) in seconds. What is the distance between a node and the next antinode?
Form: \( y = A \sin (kx) \cos (\omega t) \), \( k = \frac{\pi}{3} \, \text{rad/m} \).
Wavelength: \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{\frac{\pi}{3}} = 6 \, \text{m} \).
Distance between node and antinode: \( \frac{\lambda}{4} = \frac{6}{4} = 1.5 \, \text{m} \).
What is the primary source of restoring force for transverse waves on a stretched string?
In transverse waves on a string, the tension provides the restoring force that opposes displacement, enabling wave propagation, unlike elasticity or pressure in other contexts.
A stationary wave is given by \( y = 0.06 \sin (\frac{\pi x}{4}) \cos (120\pi t) \), where \( x \) and \( y \) are in meters and \( t \) in seconds. What is the wavelength?
Form: \( y = A \sin (kx) \cos (\omega t) \), \( k = \frac{\pi}{4} \, \text{rad/m} \).
Wavelength: \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{\frac{\pi}{4}} = 8 \, \text{m} \).
A string of length 3 m and mass 0.06 kg has a fundamental frequency of 25 Hz. What is the tension in the string?
\( \mu = \frac{0.06}{3} = 0.02 \, \text{kg/m} \).
\( v_1 = \frac{v}{2L} \Rightarrow 25 = \frac{v}{2 \times 3} \Rightarrow v = 25 \times 6 = 150 \, \text{m/s} \).
\( v = \sqrt{\frac{T}{\mu}} \Rightarrow 150 = \sqrt{\frac{T}{0.02}} \Rightarrow 150^2 = \frac{T}{0.02} \).
\( T = 22500 \times 0.02 = 450 \, \text{N} \).
A string of length 0.8 m is fixed at both ends. If the speed of the wave is 40 m/s, what is the frequency of the second harmonic?
For fixed ends: \( v_n = \frac{n v}{2L} \).
Second harmonic: \( n = 2 \).
\( v_2 = \frac{2 \times 40}{2 \times 0.8} = \frac{80}{1.6} = 50 \, \text{Hz} \).
A string of length 1.5 m and mass 0.03 kg has a wave speed of 60 m/s. What is the tension in the string?
\( \mu = \frac{0.03}{1.5} = 0.02 \, \text{kg/m} \).
\( v = \sqrt{\frac{T}{\mu}} \Rightarrow 60 = \sqrt{\frac{T}{0.02}} \Rightarrow 60^2 = \frac{T}{0.02} \).
\( T = 3600 \times 0.02 = 72 \, \text{N} \).
Why does a wave reflect with no phase change at a free end?
At a free end, the boundary is unconstrained, allowing maximum displacement. The reflected wave reinforces the incident wave in the same direction, maintaining phase.
Two strings produce beats of 3 Hz. One has a frequency of 288 Hz. When the tension in the second string is increased, the beat frequency becomes 1 Hz. What was the original frequency of the second string?
\( |288 - v_2| = 3 \Rightarrow v_2 = 285 \, \text{Hz or } 291 \, \text{Hz} \).
Increasing tension increases frequency. If \( v_2 = 285 \), new \( v_2’ > 285 \), beat = \( 288 - v_2’ < 3 \), becomes 1 Hz (\( v_2’ = 287 \)), consistent.
If \( v_2 = 291 \), beat increases, contradicts.
So, \( v_2 = 285 \, \text{Hz} \).
In a standing wave formed on a string fixed at both ends, what is the condition for the position of nodes?
In a standing wave, nodes occur where the displacement is zero, which happens when \( \sin(kx) = 0 \). This implies \( kx = n\pi \), where \( k = \frac{2\pi}{\lambda} \), so \( x = \frac{n\lambda}{2} \) (n = 0, 1, 2, ...).
Two waves \( y_1 = 3 \sin (4x - 10t) \) and \( y_2 = 3 \sin (4x - 10t + \pi/2) \) interfere. What is the amplitude of the resultant wave?
Amplitude: \( A = 2a \cos \frac{\phi}{2} \), where \( a = 3 \, \text{m} \), \( \phi = \pi/2 \).
\( A = 2 \times 3 \cos \frac{\pi/4} = 6 \times \frac{1}{\sqrt{2}} = 3\sqrt{2} \approx 4.24 \, \text{m} \).
A transverse wave on a string has a wavelength of 2 m and a period of 0.04 s. What is the speed of the wave?
Frequency: \( v = \frac{1}{T} = \frac{1}{0.04} = 25 \, \text{Hz} \).
Speed: \( v = v \lambda = 25 \times 2 = 50 \, \text{m/s} \).
A wave on a string has an amplitude of 0.03 m and a frequency of 25 Hz. What is the maximum transverse speed of a particle?
\( \omega = 2\pi v = 2\pi \times 25 = 50\pi \, \text{rad/s} \).
Max speed: \( v_{\text{max}} = a \omega = 0.03 \times 50\pi \approx 4.71 \, \text{m/s} \).
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