Question

A 25-g string is stretched with a tension of 43 N between two fixed points 12 m apart. What is the frequency of the second harmonic?

Answer 1

Answer:

The frequency of the second harmonic ($2f_o$) is 11.97 Hz.

Explanation:

Given;

mass of the string, m = 25 g = 0.025kg

tension on the string, T = 43 N

length of the string, L = 12 m

The speed of wave on the string is given as;

$v = \sqrt{\frac{T}{\mu} }$

where;

μ is mass per unit length = 0.025 / 12 = 0.002083 kg/m

$v = \sqrt{\frac{43}{0.002083} }\\\\v = 143.678 \ m/s$

The wavelength of the first harmonic wave is given as;

$L = \frac{1}{2} \lambda _o\\\\\lambda _o = 2L \\\\\lambda _o = 2 \ \times \ 12\\\\\lambda _o = 24 \ m$

The frequency of the first harmonic is given as;

$f_o = \frac{v}{\lambda _o} = \frac{v}{2L} = \frac{143.678}{24} = 5.99 \ Hz$

The wavelength of the second harmonic wave is given as;

$L = \lambda_1 \\\\\lambda_1 = 12 \ m$

The frequency of the second harmonic is given as;

$f_1 = \frac{v}{\lambda _1} = \frac{143.678}{12} = 11.97 \ Hz = 2(\frac{v}{\lambda _0}) = 2f_o$

Therefore, the frequency of the second harmonic ($2f_o$) is 11.97 Hz.