Question

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length and make it vibrate in the fundamental frequency once again. The rubber band is made so that doubling its length doubles the tension and reduces the mass per unit length by a factor of 2. The new frequency will be related to the old by a factor of:

Answer 1

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Step-by-step explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity =
`\sqrt{(T)/((M)/(L))}`

Frequency (F₁) =
`\frac{\sqrt{(T)/((M)/(L))}}{2*L}`

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ =
`\frac{\sqrt{(2T)/(0.5((M)/(L)))}}{4*L} = \frac{\sqrt{4((T)/((M)/(L))})}{4*L} = (2)/(2) [\frac{\sqrt{(T)/((M)/(L))}}{2*L}] = F_1`

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).