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?

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;

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;

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

Answer:

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters