What are (a) the lowest frequency, (b) the second lowest frequency, (c) the third lowest frequency of transverse vibrations on a wire that is 10.0m long, has a mass of 100g, and is stretched under tension of 250 N?

Answer:f1 = 7.90Hz, f2 = 15.811Hz, f3 = 23.71Hz.

Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T = 250N

We need the velocity of sound wave in a string when plucked with a tension T, this is given below as

v = √T/u

Where u = mass / length = 0.1 / 10 = 0.01kg/m

Hence v = √250/0.01, v = √25,000 = 158.11

a) at the lowest frequency.

At the lowest frequency, the length of string is related to the wavelength with the formulae below

L = λ/2, λ = 2L.

λ = 2 * 10

λ = 20m.

But v = fλ where v = 158.11m/s and λ = 20m

f = v / λ

f = 158.11 / 20

f = 7.90Hz.

b) at the first frequency.

The length of string and wavelength for this case is

L = λ.

Hence λ = 10m

v = 158.11m/s

v = fλ

f = v/λ

f = 158.11/10

f = 15.811Hz

c) at third frequency

The length of string is related to the wavelength of sound with the formulae below

L = 3λ/2, hence λ = 2L / 3

λ = 2 * 10 / 3

λ = 20/3

λ = 6.67m

v = fλ where v = 158.11m/s, λ = 6.67m

f = v/λ

f = 158.11/6.67

f = 23.71Hz.