Correct answer Carries: 4.
Wrong Answer Carries: -1.
A transverse wave on a string has a tension of 225 N and a linear mass density of 0.025 kg/m. What is the wavelength if the frequency is 30 Hz?
Speed: \( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{225}{0.025}} = \sqrt{9000} \approx 94.87 \, \text{m/s} \).
Wavelength: \( \lambda = \frac{v}{v} = \frac{94.87}{30} \approx 3.16 \, \text{m} \).
A stationary wave on a string is given by \( y = 0.08 \sin (2\pi x) \cos (100\pi t) \), where \( x \) and \( y \) are in meters and \( t \) in seconds. What is the distance between two consecutive nodes?
Form: \( y = A \sin (kx) \cos (\omega t) \), \( k = 2\pi \, \text{rad/m} \).
Wavelength: \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi} = 1 \, \text{m} \).
Distance between nodes: \( \frac{\lambda}{2} = \frac{1}{2} = 0.5 \, \text{m} \).
A stationary wave on a string fixed at both ends has a wavelength of 0.6 m and a frequency of 100 Hz. What is the wave speed?
Speed: \( v = v \lambda = 100 \times 0.6 = 60 \, \text{m/s} \).
A pipe closed at one end has a length of 0.85 m and resonates at its second harmonic with a speed of sound of 340 m/s. What is the frequency?
For closed pipe: \( v_n = (n + \frac{1}{2}) \frac{v}{2L} \), \( n = 1 \) for second harmonic.
\( v_1 = (1 + \frac{1}{2}) \frac{340}{2 \times 0.85} = 1.5 \times \frac{340}{1.7} = 300 \, \text{Hz} \).
A string of length 2.5 m and mass 0.025 kg has a fundamental frequency of 40 Hz. What is the tension in the string?
\( \mu = \frac{0.025}{2.5} = 0.01 \, \text{kg/m} \).
\( v_1 = \frac{v}{2L} \Rightarrow 40 = \frac{v}{2 \times 2.5} \Rightarrow v = 40 \times 5 = 200 \, \text{m/s} \).
\( v = \sqrt{\frac{T}{\mu}} \Rightarrow 200 = \sqrt{\frac{T}{0.01}} \Rightarrow 200^2 = \frac{T}{0.01} \).
\( T = 40000 \times 0.01 = 400 \, \text{N} \).
A string of length 2.4 m fixed at both ends has a wave speed of 72 m/s. What is the frequency of its fifth harmonic?
\( v_n = \frac{n v}{2L} \).
Fifth harmonic (\( n = 5 \)): \( v_5 = \frac{5 \times 72}{2 \times 2.4} = \frac{360}{4.8} = 75 \, \text{Hz} \).
Two strings produce beats of 8 Hz. One has a frequency of 440 Hz. When the tension in the second string is increased, the beat frequency becomes 6 Hz. What was the original frequency of the second string?
Let \( v_2 \) be the original frequency.
\( |440 - v_2| = 8 \Rightarrow v_2 = 432 \, \text{Hz or } 448 \, \text{Hz} \).
Increasing tension increases frequency. If \( v_2 = 432 \), new \( v_2’ > 432 \), beat = \( 440 - v_2’ < 8 \), becomes 6 Hz (\( v_2’ = 434 \)), consistent.
If \( v_2 = 448 \), beat increases, contradicts.
So, \( v_2 = 432 \, \text{Hz} \).
Two waves \( y_1 = 3 \sin (10x - 20t) \) and \( y_2 = 3 \sin (10x - 20t + \frac{\pi}{4}) \) interfere. What is the amplitude of the resultant wave?
Amplitude: \( A = 2a \cos \frac{\phi}{2} \), \( a = 3 \, \text{m} \), \( \phi = \frac{\pi}{4} \).
\( A = 2 \times 3 \cos \frac{\pi}{8} \approx 6 \times 0.9239 \approx 5.54 \, \text{m} \).
A pipe 0.5 m long, open at both ends, resonates with a 340 Hz source. What is the harmonic number if the speed of sound is 340 m/s?
For open pipe: \( v_n = \frac{n v}{2L} \).
Given: \( v = 340 \, \text{m/s} \), \( L = 0.5 \, \text{m} \), \( v_n = 340 \, \text{Hz} \).
\( 340 = \frac{n \times 340}{2 \times 0.5} \Rightarrow 340 = 340n \Rightarrow n = 1 \).
First harmonic (fundamental mode).
A pipe closed at one end has a length of 0.25 m. What is the frequency of its third harmonic if the speed of sound is 340 m/s?
For closed pipe: \( v_n = (n + \frac{1}{2}) \frac{v}{2L} \), \( n = 0, 1, 2, \ldots \).
Third harmonic: \( n = 2 \).
\( v_2 = (2 + \frac{1}{2}) \frac{340}{2 \times 0.25} = 2.5 \times \frac{340}{0.5} = 1700 \, \text{Hz} \).
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