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Chapter 13: Problem 64

A certain tuning fork vibrates at a frequency of \(196 \mathrm{~Hz}\) while each tip of its two prongs has an amplitude of \(0.850 \mathrm{~mm}\). (a) What is the period of this motion? (b) Find the wavelength of the sound produced by the vibrating fork, taking the speed of sound in air to be \(343 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The period of this motion is 0.00510 s and the wavelength of the sound produced by the vibrating fork is 1.75 m.

Step by step solution

01

Calculate the Period

The period of the motion is the reciprocal of the frequency, which in this case means \(T = 1/f\). Thus, \(T = 1 / 196 Hz = 0.00510 s\).
02

Find the Wavelength

For finding the wavelength, the formula speed = wavelength * frequency is used. Therefore, rearranging the formula to find the wavelength gives \(wavelength = speed/frequency\), substituting the given values will give: \(wavelength = 343 m/s / 196 Hz = 1.75 m\).

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