Correct answer Carries: 4.
Wrong Answer Carries: -1.
A pipe open at both ends has a length of 0.425 m and resonates with a source of frequency 800 Hz. What is the harmonic number if the speed of sound is 340 m/s?
\( v_n = \frac{n v}{2L} \).
\( 800 = \frac{n \times 340}{2 \times 0.425} = \frac{n \times 340}{0.85} \).
\( 800 = n \times 400 \Rightarrow n = \frac{800}{400} = 2 \).
Second harmonic.
A pipe closed at one end has a length of 0.45 m and a speed of sound of 360 m/s. What is the frequency of its fifth harmonic?
For closed pipe: \( v_n = (n + \frac{1}{2}) \frac{v}{2L} \), \( n = 4 \) for fifth harmonic.
\( v_4 = (4 + \frac{1}{2}) \frac{360}{2 \times 0.45} = 4.5 \times \frac{360}{0.9} = 4.5 \times 400 = 1800 \, \text{Hz} \).
A longitudinal wave in a medium has a speed of 1600 m/s and a bulk modulus of 2.56 × 10⁹ Pa. What is the density of the medium?
Speed: \( v = \sqrt{\frac{B}{\rho}} \).
\( 1600 = \sqrt{\frac{2.56 \times 10^9}{\rho}} \Rightarrow 1600^2 = \frac{2.56 \times 10^9}{\rho} \).
\( \rho = \frac{2.56 \times 10^9}{1600^2} = \frac{2.56 \times 10^9}{2.56 \times 10^6} = 1000 \, \text{kg/m}^3 \).
Two strings produce beats of 5 Hz. One has a frequency of 256 Hz. When the tension in the second string is decreased, the beat frequency becomes 3 Hz. What was the original frequency of the second string?
Let \( v_2 \) be the original frequency.
\( |256 - v_2| = 5 \Rightarrow v_2 = 251 \, \text{Hz or } 261 \, \text{Hz} \).
Decreasing tension decreases frequency. If \( v_2 = 261 \), new \( v_2’ < 261 \), beat = \( 261 - 256 = 5 \) or decreases, so \( v_2’ = 253 \), beat = \( 256 - 253 = 3 \), consistent.
Thus, \( v_2 = 261 \, \text{Hz} \).
Two sitar strings produce beats of 5 Hz. If one has a frequency of 320 Hz and tension in the other is reduced, the beat frequency increases to 7 Hz. What is the original frequency of the second string?
\( |320 - v_2| = 5 \Rightarrow v_2 = 315 \, \text{Hz or } 325 \, \text{Hz} \).
Reducing tension decreases frequency. If \( v_2 = 325 \), new \( v_2’ < 325 \), beat = \( 325 - 320 = 5 \) or increases, so \( v_2’ = 313 \), beat = \( 320 - 313 = 7 \), consistent.
Thus, \( v_2 = 325 \, \text{Hz} \).
A pipe open at both ends resonates at 510 Hz with a speed of sound of 340 m/s. What is its length?
Fundamental: \( v_1 = \frac{v}{2L} \).
\( 510 = \frac{340}{2L} \Rightarrow 2L = \frac{340}{510} \Rightarrow 2L = \frac{2}{3} \Rightarrow L = 0.333 \, \text{m} \).
A string of length 2.8 m and mass 0.07 kg has a fundamental frequency of 35 Hz. What is the tension in the string?
\( \mu = \frac{0.07}{2.8} = 0.025 \, \text{kg/m} \).
\( v_1 = \frac{v}{2L} \Rightarrow 35 = \frac{v}{2 \times 2.8} \Rightarrow v = 35 \times 5.6 = 196 \, \text{m/s} \).
\( v = \sqrt{\frac{T}{\mu}} \Rightarrow 196 = \sqrt{\frac{T}{0.025}} \Rightarrow 196^2 = \frac{T}{0.025} \).
\( T = 38416 \times 0.025 = 960.4 \, \text{N} \).
Which property of a sound wave is most directly responsible for its loudness?
Loudness is primarily determined by the amplitude of a sound wave, as it relates to the energy carried and perceived intensity, while frequency affects pitch.
A pipe closed at one end has a length of 0.8 m and resonates at its third harmonic with a speed of sound of 360 m/s. What is the frequency?
For closed pipe: \( v_n = (n + \frac{1}{2}) \frac{v}{2L} \), \( n = 2 \) for third harmonic.
\( v_2 = (2 + \frac{1}{2}) \frac{360}{2 \times 0.8} = 2.5 \times \frac{360}{1.6} = 2.5 \times 225 = 562.5 \, \text{Hz} \).
Which type of wave cannot propagate through a vacuum?
Mechanical waves, such as sound (longitudinal) waves, require a medium for propagation because they rely on particle vibrations. Electromagnetic waves, like light, can travel through a vacuum.
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