JAMB 2011 Physics · Question 32

A length of wire has a frequency of 225Hz when stretched by a force of 225N. If the force increased to 324N, what is the new frequency of vibration?

Options

A.270 Hz
B.512 Hz
C.488 Hz
D.356 Hz

Answer: A. 270 Hz

Explanation written with AI and checked against the answer key

Analyze the question A stretched wire (string) vibrates at a frequency that depends on the tension. The length and mass per unit length stay the same, so only the tension changes. Concept For a vibrating string, the frequency is: f = (1/2L) √(T/μ) So f ∝ √T when L and μ are constant. This gives: f₂/f₁ = √(T₂/T₁) Solution Given: f₁ = 225 Hz, T₁ = 225 N, T₂ = 324 N f₂ = f₁ × √(T₂/T₁) f₂ = 225 × √(324/225) √324 = 18 and √225 = 15, so: f₂ = 225 × (18/15) f₂ = 225 × 1.2 f₂ = 270 Hz Answer The new frequency is 270 Hz, which is option A. Tip Look for perfect squares. When numbers like 225 and 324 appear, they are usually chosen so the square root works out cleanly (15 and 18). Also, if the tension goes up, the frequency must go up, but only by the square root of the ratio. That lets you eliminate B and C, which are far too large, and D, which is too big for a tension increase of only about 44%.

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