AS Daily A Level Physics question
In a lab sonometer, a steel string of length L is fixed at both ends and under tension T. It is driven at a clear third-harmonic resonance with frequency f3. A light clamp is then added exactly at the midpoint to hold that point stationary, and at the same time the tension is increased to 2T. The driver is swept to find the lowest-frequency resonance of the modified setup. Compared with the original f3, what is the new resonance frequency, and why?
Answer
The correct answer is A.
Correct: A — It decreases to about 0.94 f3, because the clamp forces a node at the centre so the lowest allowed mode is n = 2, and doubling T increases the wave speed by a factor √2. A is correct since only even modes survive when the midpoint is fixed (lowest n = 2) and f ∝ n × wave speed, with the speed scaling as √T, giving f_new/f3 = (2√2)/3 ≈ 0.94. B assumes the wave speed is proportional to T (so doubling T would double the speed), leading to a ratio of about 4/3 ≈ 1.33, which ignores the √T dependence. C ignores the change in tension, taking only the harmonic-number change (2/3) and missing the √2 increase from the higher tension. D incorrectly claims little or no change: the third harmonic has an antinode at the centre (so it is disallowed when clamped), and the new lowest mode is n = 2 with frequency v′/L, which is not equal to the old f3.