Correct answer Carries: 4.
Wrong Answer Carries: -1.
A string fixed at both ends has a length of 1.5 m and a fundamental frequency of 50 Hz. What is the speed of the wave on the string?
Fundamental frequency: \( v_1 = \frac{v}{2L} \).
\( 50 = \frac{v}{2 \times 1.5} \Rightarrow 50 = \frac{v}{3} \Rightarrow v = 150 \, \text{m/s} \).
Which wave property is most affected when a sound wave encounters a change in the medium’s temperature?
Speed of sound in a gas increases with temperature (\( v \propto \sqrt{T} \)), significantly altering its propagation, while frequency remains source-dependent and amplitude may vary less directly.
Why are standing waves formed in musical instruments like a guitar?
Standing waves in a guitar arise from the interference of waves reflected at fixed ends (e.g., bridge and nut), producing discrete frequencies (harmonics) determined by string length.
A string fixed at both ends has a length of 1.4 m and a wave speed of 70 m/s. What is the frequency of its second harmonic?
For fixed ends: \( v_n = \frac{n v}{2L} \).
Second harmonic (\( n = 2 \)): \( v_2 = \frac{2 \times 70}{2 \times 1.4} = \frac{140}{2.8} = 50 \, \text{Hz} \).
Two strings produce beats of 4 Hz. One has a frequency of 320 Hz. When the tension in the second string is increased, the beat frequency becomes 2 Hz. What was the original frequency of the second string?
Let \( v_2 \) be the original frequency.
\( |320 - v_2| = 4 \Rightarrow v_2 = 316 \, \text{Hz or } 324 \, \text{Hz} \).
Increasing tension increases frequency. If \( v_2 = 316 \), new \( v_2’ > 316 \), beat = \( 320 - v_2’ < 4 \), becomes 2 Hz (\( v_2’ = 318 \)), consistent.
If \( v_2 = 324 \), beat increases, contradicts.
So, \( v_2 = 316 \, \text{Hz} \).
A string of length 2.2 m fixed at both ends has a wave speed of 66 m/s. What is the frequency of its fourth harmonic?
\( v_n = \frac{n v}{2L} \).
Fourth harmonic (\( n = 4 \)): \( v_4 = \frac{4 \times 66}{2 \times 2.2} = \frac{264}{4.4} = 60 \, \text{Hz} \).
A string of length 2 m and mass 0.01 kg is under a tension of 100 N. What is the time taken by a transverse pulse to travel from one end to the other?
Linear mass density: \( \mu = \frac{\text{mass}}{\text{length}} = \frac{0.01}{2} = 0.005 \, \text{kg/m} \).
Speed: \( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.005}} = \sqrt{20000} \approx 141.4 \, \text{m/s} \).
Time: \( t = \frac{\text{length}}{v} = \frac{2}{141.4} \approx 0.014 \, \text{s} \).
Two waves \( y_1 = 3 \sin (8x - 16t) \) and \( y_2 = 3 \sin (8x - 16t + \frac{2\pi}{3}) \) interfere. What is the amplitude of the resultant wave?
Amplitude: \( A = 2a \cos \frac{\phi}{2} \), \( a = 3 \, \text{m} \), \( \phi = \frac{2\pi}{3} \).
\( A = 2 \times 3 \cos \frac{\pi}{3} = 6 \times \frac{1}{2} = 3 \, \text{m} \).
A string fixed at both ends has a length of 1.6 m and a fundamental frequency of 62.5 Hz. What is the speed of the wave?
Fundamental: \( v_1 = \frac{v}{2L} \).
\( 62.5 = \frac{v}{2 \times 1.6} \Rightarrow v = 62.5 \times 3.2 = 200 \, \text{m/s} \).
A steel rod of length 2 m has a fundamental frequency of longitudinal vibrations of 1.25 kHz. What is the speed of sound in the rod?
For rod clamped at middle, fundamental: \( v_1 = \frac{v}{2L} \).
\( 1250 = \frac{v}{2 \times 2} \Rightarrow v = 1250 \times 4 = 5000 \, \text{m/s} \).
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