Correct answer Carries: 4.
Wrong Answer Carries: -1.
What is the effect on a standing wave’s frequency if the length of the vibrating medium is halved?
For a standing wave, \( v_n = \frac{n v}{2L} \). Halving the length \( L \) to \( L/2 \) gives \( v_n' = \frac{n v}{2 \cdot (L/2)} = \frac{n v}{L} = 2 \cdot \frac{n v}{2L} \), doubling the frequency.
A pipe open at both ends has a length of 0.8 m and resonates with a source of frequency 637.5 Hz. What is the harmonic number if the speed of sound is 340 m/s?
\( v_n = \frac{n v}{2L} \).
\( 637.5 = \frac{n \times 340}{2 \times 0.8} = \frac{n \times 340}{1.6} \).
\( 637.5 = n \times 212.5 \Rightarrow n = \frac{637.5}{212.5} = 3 \).
Third harmonic.
What is the key requirement for a medium to support both transverse and longitudinal waves?
A medium must be rigid (have shear modulus) to support transverse waves and compressible (have bulk modulus) for longitudinal waves, properties typical of solids.
Two waves of frequencies 490 Hz and 495 Hz interfere. How many beats are heard in 8 seconds?
Beat frequency: \( v_{\text{beat}} = 495 - 490 = 5 \, \text{Hz} \).
Beats in 8 s: \( 5 \times 8 = 40 \).
Which property of a medium allows longitudinal waves to propagate through it but not transverse waves?
Longitudinal waves require a medium with bulk modulus (resistance to compression), which fluids possess. Transverse waves need shear modulus (resistance to shearing), which fluids lack due to their inability to sustain shear stress.
A wave on a string has an amplitude of 0.035 m and a frequency of 20 Hz. What is the maximum transverse speed of a particle?
\( \omega = 2\pi v = 2\pi \times 20 = 40\pi \, \text{rad/s} \).
Max speed: \( v_{\text{max}} = a \omega = 0.035 \times 40\pi \approx 4.4 \, \text{m/s} \).
A wave is described by \( y(x, t) = 0.02 \sin (25x - 75t) \), where \( x \) and \( y \) are in meters and \( t \) in seconds. What is the period of the wave?
Compare with \( y = a \sin (kx - \omega t) \), \( \omega = 75 \, \text{rad/s} \).
Period: \( T = \frac{2\pi}{\omega} = \frac{2\pi}{75} \approx 0.0838 \, \text{s} \).
Two strings produce beats of 3 Hz. One has a frequency of 400 Hz. When the tension in the second string is slightly increased, the beat frequency becomes 2 Hz. What was the original frequency of the second string?
Let \( v_2 \) be the original frequency.
\( |400 - v_2| = 3 \Rightarrow v_2 = 397 \, \text{Hz or } 403 \, \text{Hz} \).
Increasing tension increases frequency. If \( v_2 = 397 \), new \( v_2’ > 397 \), beat = \( 400 - v_2’ < 3 \), becomes 2 Hz, consistent (\( v_2’ = 398 \)).
If \( v_2 = 403 \), beat increases, contradicts.
So, \( v_2 = 397 \, \text{Hz} \).
A wave \( y = 0.03 \sin (20x - 40t) \) reflects at an open boundary. What is the equation of the reflected wave?
At open boundary, no phase change.
Incident: \( y_i = 0.03 \sin (20x - 40t) \).
Reflected: \( y_r = 0.03 \sin (20x + 40t) \).
Two waves of frequencies 496 Hz and 502 Hz interfere. How many beats are heard in 10 seconds?
Beat frequency: \( v_{\text{beat}} = 502 - 496 = 6 \, \text{Hz} \).
Beats in 10 s: \( 6 \times 10 = 60 \).
Are you sure you want to submit your answers?
