PHYSICS

JAMB 2021 - Question 21

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?

Choose the correct answers from the options given.
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?
A:
B:
C:
D:
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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.

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