Question

You haw a uniform string with a mass of 0.0130 kg and length 1.75 m under a tension 10.0 N. The string is fixed at both ends, and is vibrating at its fourth resonant frequency (i.e. the fourth harmonic). What is the wavelength of the standing wave in the string? What is the frequency?

Answer 1

Velocity ven 0 Co 0135/1.ts) m /L、 = 367 mls 2 L 2x).75 1.93再 CA,s) 41.93

Answer 2

Wavelength

For a standing transverse wave in a stretched string that is attached at both ends, the number of the normal mode equals the number of half wavelengths that are contained in its length. The wavelength ??${\mathit{\lambda }}_n$ of the ?$n$th normal mode is therefore

$${\mathit{\lambda }}_n=\frac{2L}{n}n=1,2,3,...$$

where ?$L$ denotes the string's length.

In the question, you are given ?$L$ and ?$n$. Substitute to obtain the numerical answer for the wavelength.

$${\mathit{\lambda }}_4=\frac{2\left(1.91\text{ m}\right)}{4}=0.955\text{ m}$$

Frequency

The frequency of the standing wave in the ?$n$th normal mode, which is the ?$n$th harmonic ??$f_n$, can be found using the relation that is valid for all periodic waves

$$f_n=\frac{v}{{\mathit{\lambda }}_n}$$

where ??${\mathit{\lambda }}_n$ is the wavelength of the ?$n$th normal mode (see the formula for wavelength) and ?$v$ is the speed of transverse waves in the string. This speed is not given directly, but can be obtained from the formula

$$v=\sqrt{\frac{F}{m/L}}$$

where ?$F$ is the given tension in the string and ?$m$ and ?$L$ are its given mass and length, respectively. Perform all necessary substitutions to obtain the frequency ?4$f_4$ in terms of given quantities.

$$f_n=\frac{n}{2L}\sqrt{\frac{F}{m/L}}n=1,2,3,...$$

For the value of the fourth harmonic,

?4=42(1.91 m)(8.17 N)(0.0135 kg)/(1.91 m)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=35.6 Hz