# Amplitude: Speed of sound waves in water, acoustic intensity, pulses, incident energy, piano note frequencies, isotropic point sources of sound

1. The speed of sound waves in water (density 1.00 x 10 3 kg/m 3 ) is c=1480 m/s.

[2] (a) What is the average intensity (in W/m 2 ) of a 1000 Hz sound wave in water if the displacement amplitude of the wave is 1.00 nm? (1 nm=10 -9 m).

[2] (b) What is the ratio of pressure amplitude to displacement amplitude in this wave?

[2] (c) Assuming an absorption coefficient of 0.0001 dB/m in the water, how far does a plane wave have to travel in the water so that its displacement amplitude is reduced from 1.0nm to 0.1 nm?

2. Two isotropic acoustic sources are located a distance of 10.0 m apart along an x-axis. The source at x=0 puts out a total acoustic power of 1.00 mW, while the source at x=10 m puts out a total power of 2.00 mW. Assuming that the acoustic intensity at any x between the sources is just the sum of the intensities arriving from the two sources,[3] (a) at what value of x between the sources is the acoustic intensity a minimum (ignore any effects due to absorption in the surrounding air)?

[3] (b) What is the minimum intensity between the sources, and what is the corresponding intensity level in dB?

[6] 3. A string of mass per unit length ??is joined to a string of mass per unit length 1.5 ??at x=0

in such a way that the tensions in the two strings are different. (This can be done by using a mass less, frictionless slip ring as illustrated in Figure 5.4 in M10-5, p. 97.) The lighter string is under twice the tension of the heavier string. If a transverse wave pulse of amplitude A is incident in the lighter string,

(i) What are the amplitudes of the pulses reflected and transmitted at the junction between the two strings?

(ii) What fraction of the incident energy is transmitted?

[3] 4.(a) The frequency of the note referred to as "middle C" on the piano is approximately 256!Hz. Suppose the string that oscillates at 256 Hz in its fundamental mode is 1.00 m long and has mass per unit length 2.00 g/cm. What tension must be applied to this string to produce the desired frequency?

[3] (b) The lowest note on the piano corresponds to a frequency of 32 Hz. Assuming that the tension applied to this string is the same as that found for "middle C" in part (a), and that the string must have length 2.00 m, what is the required mass per unit length?

5. Two isotropic point sources of sound are locate distances of 5.00 m and 3.00 m away

from a point P in a lossless (i.e., non-absorbing) medium in which the sound speed is 340!m/s. The sources operate at identical frequencies of 1000 Hz, and the power emitted by each source is equal to 5.27 mW. The waves from the two sources are traveling in very nearly the same direction when they reach P.

[4] (a) The phases of the simple harmonic vibrations of the sources are in phase with each other.

Assuming coherent superposition, find the resultant acoustic intensity and the intensity level at point P. (Remember that the amplitudes at P are not identical.)

[2] (b) The phase of the more distant source (the one 5.00 m away from P) is now slowly increased until an interference maximum (i.e., constructive interference) is achieved at P.

By how many radians has the phase been increased when constructive interference is first achieved? (Increasing the phase of a source causes a given feature, such as a wave crest, to be emitted earlier in time.) Also find the value of the intensity at P under the new

circumstances.

#### Solution Preview

To solve this question, we need to understand few concepts of sound waves and their properties. I assume you have encountered these before in your class, so I will just list the formulae.

For a sound wave, we can describe it in terms of its displacement as :

s(x,t)= smax*cos(kx-wt),

or in terms of its pressure as :

dP(x,t)=dPmax*cos(kx-wt)

Here smax = Maximum displacement amplitude, dPmax = Maximum pressure amplitude, w=angular frequency of the wave.

(i) The displacement amplitude and pressure ...

#### Solution Summary

With excellent explanations and calculations, the problems are solved in the following response.