Question
A guitar string vibrates in a manner resulting in a standing wave having a wavelength $\lambda$ being formed in the string. A listener hears the fundamental frequency, $f_{1},$ for this particular string. The string is plucked a second time in a manner that roduces the second harmonic for this string. How does the wavelength for the second harmonic compare with the wavelength at the fundemental frequency?(A) $\frac{1}{2} \lambda$(B) $\frac{\sqrt{2}}{2} \lambda$(C) $\lambda$(D) $\sqrt{2} \lambda$(E) 2$\lambda$
Step 1
The length of the string, $L$, is half the wavelength, $\lambda$, of the wave. This can be written as: \[L = \frac{\lambda}{2}\] or \[\lambda = 2L\] Show more…
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A guitar string vibrates in a manner resulting in a standing wave having a wavelength $\lambda$ being formed in the string. A listener hears the fundamental frequency, $f_{1},$ for this particular string. The string is plucked a second time in a manner that produces the second harmonic for this string. How does the wavelength for the second harmonic compare with the wavelength at the fundamental frequency? (A) $\frac{1}{2} \lambda$ (B) $\frac{\sqrt{2}}{2} \lambda$ (C) $\lambda$ (D) $\sqrt{2} \lambda$ (E) 2$\lambda$
A guitar string oscillates at a frequency of $100 \mathrm{Hz}$ and produces a sound wave. (a) What do you think the frequency of the sound wave is that the vibrating string produces? (b) If the speed of the sound wave is $v=343 \mathrm{m} / \mathrm{s},,$ what is the wavelength of the sound wave?
Both of the above strings have their ends locked in place. The first string, $S_{1},$ is twice as long as the second string, $S_{2}$ . If sound waves are going to be sent through both, what is the correct ratio of the fundamental frequency of $S_{1}$ the fundamental frequency of $S_{2} ?$ (A) $2 : 1$ (B) $\sqrt{2} : 1$ (B) $\sqrt{2} : 1$ (C) $1 : \sqrt{2}$ (D) $1 : 2$
Practice Test 1
Section 1
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