Question

A string of density 0.01 kg/m is stretched with a tension of 5N and fixed at both ends. The length of the string is 0.1m. What is the first four resonance frequencies in the string?

Answer 1

Answer:

The first four resonance frequency in the string are;

1) 50·√50 Hz

2) 100·√50 Hz

3)150·√50 Hz

4) 200·√50 Hz

Explanation:

The given parameters of the string are;

The density of the string, ρ = 0.01 kg/m

The tension force on the string, T = 5 N

The length of the string, l = 0.1 m

Therefore the mass of the string, m = Length of string × Density of the string

∴ m = 0.01 kg/m × 0.1 m = 0.001 kg

The formula for the fundamental frequency, f₁, is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{\sqrt{\dfrac{T}{\rho} } }{2 \cdot L}$

Where;

f₁ = The fundamental frequency in the string

T = The tension in the string = 5 N

m = The mass of the string = 0.001 kg

L = The length of the string = 0.1 m

ρ = The density of the string = 0.01 kg/m

By plugging in the values of the variables, we have;

$f_1 = \dfrac{\sqrt{\dfrac{5}{0.01} } }{2 \times 0.1} = 50 \cdot \sqrt{5}$

The first four harmonics are;

f₁, 2·f₁, 3·f₁, 4·f₁

Therefore, we have the first four resonance frequency of the string are as follows;

1 × 50·√50 Hz = 50·√50 Hz

2 × 50·√50 Hz = 100·√50 Hz

3 × 50·√50 Hz = 150·√50 Hz

4 × 50·√50 Hz  = 200·√50 Hz