Question

1. In a demonstration, Mr. K stretches a steel wire to a length 1.23 meters and braces both ends so that they are not free to vibrate. He attaches a fancy piece of equipment which he calls a mechanical oscillator to the wire and explains how it works. Then Mr. K turns the oscillator on and tunes the frequency to 588 Hz. To the amazement of the class, the wire begins vibrating in the sixth harmonic wave pattern. a. Determine the speed of waves within the wire. b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern. c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.

Answer 1

Answer:

a.241.08 m/s b. 196 Hz c. 392 Hz

Explanation:

a. Determine the speed of waves within the wire.

The frequency of oscillation of the wave in the string, f = nv/2L where n = harmonic number, v = speed of wave in string, L = length of string = 1.23 m.

Since f = 588 Hz which is the 6 th harmonic, n = 6. So, making v subject of the formula, we have

v = 2Lf/n

substituting the values of the variables into v. we have

v = 2 × 1.23 m × 588Hz/6

v = 241.08 m/s

b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern.

The first harmonic is obtained from f when n = 1,

So, f = v/2L = 241.08 m/s ÷ 1.23m = 196 Hz

c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.

The second harmonic f' = 2f = 2 × 196 Hz = 392 Hz