Question

1. a string with a mass per unit length of 1.5 g/m is put under a tension of 20 n. what length is required for its fundamental resonance frequency to be 220 hz?

a. 0.26 m
b. 0.96 m
c. 1.1 m
d. 1.2 m

1. for a periodic wave, the displacement from the equilibrium point to the crest/trough of the wave is called the

a. wavelength
b. amplitude
c. frequency
d. period

1. find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when it is subjected to tension of 50.0 n.

a. 132 hz, 264 hz and 396 hz
b. 66 hz, 132 hz and 198 hz
c. 264 hz, 528 hz and 792 hz
d. none of the given

Explanation:

Step1: Calculate the wave speed \(v\)

The wave speed \(v\) on a string is given by \(v = \sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the mass - per - unit - length.
Given \(T = 50\space N\) and \(\mu=2.00\times10^{- 3}\space kg/m\) (since \(2.00\space g/m=2.00\times10^{-3}\space kg/m\))

$$v=\sqrt{\frac{50}{2.00\times 10^{-3}}}=\sqrt{25000}=158.11\space m/s$$

Step2: Calculate the fundamental frequency \(f_1\)

For a string fixed at both ends, the fundamental frequency \(f_1=\frac{v}{2L}\), where \(L = 0.600\space m\)

$$f_1=\frac{158.11}{2\times0.600}=131.76\space Hz\approx132\space Hz$$

Step3: Calculate the second harmonic \(f_2\)

The \(n\)th harmonic \(f_n=nf_1\). For \(n = 2\), \(f_2 = 2f_1\)

$$f_2=2\times132 = 264\space Hz$$

Step4: Calculate the third harmonic \(f_3\)

For \(n = 3\), \(f_3=3f_1\)

$$f_3=3\times132=396\space Hz$$

Answer:

A. \(132\space Hz,264\space Hz\) and \(396\space Hz\)