Physics 2021 JAMB Past Questions - Question 21: A vibrating spring has a tension of 40N and produced a note of 200Hz when picked in the middle, the length of the string is unaltered and the tension is increased by 120N, then the new frequency becomes?

Correct Answer

D

Explanation

To find the new frequency when the tension is increased by 120N, we can use the formula for the frequency of a vibrating spring:

f = (1/2L) * sqrt(T/μ)

Where:
f = frequency
L = length of the string
T = tension
μ = linear mass density (mass per unit length)

Given:
Initial tension, T1 = 40N
Initial frequency, f1 = 200Hz
Change in tension, ΔT = 120N

First, we can find the linear mass density (μ) using the initial tension and frequency:

f1 = (1/2L) * sqrt(T1/μ)

Solving for μ:
μ = T1 / (4 * (f1^2) * (L^2))

Now, we can find the new frequency (f2) when the tension is increased by 120N:

f2 = (1/2L) * sqrt((T1 + ΔT)/μ)

Substitute the value of μ and solve for f2:
f2 = (1/2L) * sqrt((T1 + ΔT) / (T1 / (4 * (f1^2) * (L^2))))

f2 = f1 * sqrt((T1 + ΔT) / T1)

f2 = 200Hz * sqrt((40N + 120N) / 40N)
f2 = 200Hz * sqrt(160N / 40N)
f2 = 200Hz * sqrt(4)
f2 = 200Hz * 2
f2 = 400Hz

Therefore, when the tension is increased by 120N, the new frequency becomes 400Hz.