Question

Find the frequencies of the first three harmonics of a 1.0-m long string which has a mass per unit length of 2.0  10–3 kg/m and a tension of 80 N when both ends are fixed in place.

Answer 1

Answer:

$f_1 = 100Hz$

$f_2 = 200Hz$

$f_3 = 300Hz$

Explanation:

Given:

Length of the string, L = 1 m

Mass per unit length, (m/L) = 2.0 × 10⁻³ kg/m

Tension in the string, T = 80N

Now, We know that,

Frequency, $f_n= n\frac{V}{2L}$  ................(1)

where, V = velocity

also,

$V=\sqrt{\frac{T}{m/L}}$

substituting the values in the equation we get

$V=\sqrt{\frac{80N}{2\times10^{-3}}}$

$V=200 m/s$

Now using the equation (1)

$f_1= \frac{200}{2\times 1}=100 Hz$

also,

$f_2= 2\times \frac{200}{2\times 1}=200 Hz$

$f_3= 3\times \frac{200}{2\times 1}=300 Hz$

Hence, the required frequencies are

$f_1 = 100Hz$

$f_2 = 200Hz$

$f_3 = 300Hz$