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question 20 (1 point) a string fixed at both ends is vibrating in one o…

Question

question 20 (1 point)
a string fixed at both ends is vibrating in one of its harmonics. if we now increase only the frequency at which the string is vibrating, which of the following characteristics can be increased?

a) the amplitude of the traveling waves on the string

b) the wavelength of the traveling waves on the string

c) the period of the traveling waves on the string

d) the speed of the traveling waves on the string

e) none of the above

Explanation:

Brief Explanations
  1. For a string fixed at both ends, the wave speed \( v \) depends on tension \( T \) and linear mass density \( \mu \) (\( v = \sqrt{\frac{T}{\mu}} \)), so it's constant if \( T \) and \( \mu \) don't change. Increasing frequency doesn't affect speed (eliminates d).
  2. The relationship \( v = f\lambda \) (where \( f \) is frequency, \( \lambda \) is wavelength) means if \( v \) is constant and \( f \) increases, \( \lambda \) must decrease (eliminates b).
  3. Period \( T = \frac{1}{f} \), so increasing \( f \) decreases \( T \) (eliminates c).
  4. Amplitude is related to energy input. Increasing frequency alone (without changing energy) doesn't increase amplitude, but if we consider that to increase frequency (by, e.g., driving the string harder at a higher frequency), amplitude could increase. However, the other options are definitively wrong. Wait, no—wait, the key: the string is vibrating in a harmonic. The wave speed is fixed by \( T \) and \( \mu \). Frequency \( f \) and wavelength \( \lambda \) are related by \( v = f\lambda \), so \( \lambda = v/f \). So if \( f \) increases, \( \lambda \) decreases. Period \( T = 1/f \), so \( T \) decreases. Speed \( v \) is constant. Amplitude: amplitude of standing waves (but the question says "traveling waves"—wait, on a string fixed at both ends, the waves are standing, but composed of traveling waves. However, amplitude of the traveling waves: amplitude is independent of frequency (depends on energy). But the other options are all wrong. Wait, the options: a) amplitude—can amplitude be increased by increasing frequency? If you drive the string at a higher frequency with more energy, yes, but the question says "increase only the frequency". If "only" means keeping energy same, amplitude doesn't change. But the other options are definitely wrong. So the correct answer is e? Wait no, wait: let's re-examine.

Wait, the problem says "a string fixed at both ends is vibrating in one of its harmonics. If we now increase only the frequency at which the string is vibrating..."

  • Wave speed \( v = \sqrt{T/\mu} \): depends on tension and mass per unit length, so constant. So d is out.
  • \( v = f\lambda \), so \( \lambda = v/f \). If \( f \) increases, \( \lambda \) decreases. So b is out.
  • Period \( T = 1/f \), so \( T \) decreases. So c is out.
  • Amplitude: amplitude of the standing wave (or the traveling waves) is related to the energy. If we "increase only the frequency" (i.e., change the driving frequency to a higher one, without changing the driving amplitude/energy), then the amplitude of the resulting vibration (if it's a resonant harmonic) would depend on resonance. But if we are not at resonance, but the question says "vibrating in one of its harmonics"—so it's at resonance. So to increase the frequency, we have to go to a higher harmonic (since harmonics have \( f_n = nf_1 \)). In a higher harmonic, the amplitude can be different, but the question is "which of the following can be increased". The other options are impossible. So the only possible is a? Wait, no—wait, maybe I made a mistake. Let's check again:

Wait, the key is that the four options:

a) amplitude of traveling waves: can amplitude increase? If you drive the string at a higher frequency with more power, yes. But the question says "increase only the frequency"—does "only" mean keeping all other factors (like driving amplitude) the same? If so, amplitude doesn't change. But the other options are all wrong. So the correct answer is e) none of the above? But that seems odd. Wait, no—wait, maybe the answer is a. Wait, no, let'…

Answer:

e) none of the above