Beats from a Stretched Wire Under Variable Tension
A steel wire 0.65 m long and of linear mass density 4.0 g/m is stretched between two fixed bridges. Its tension is 106 N. When the wire is plucked, it sounds its fundamental mode together with a tuning fork marked 128 Hz. A student hears 3 beats per second. She then increases the tension very slightly; the beat frequency decreases, reaches zero, and then increases again as she keeps increasing the tension. The wire is still in its fundamental mode each time. Assume the bridges do not move and the wire behaves as an ideal string. The tuning-fork frequency is known to within 0.5 Hz. She records the tension at which the beats vanish and wants to use it to check the tuning-fork frequency.
1. Calculate the fundamental frequency of the wire at 106 N. Use this and the beat count to decide whether the wire’s frequency is above or below 128 Hz.
Read the worked answer
Wave speed v = √(T/μ) = √(106/(4.0 × 10⁻³)) = √(2.65 × 10⁴) = 163 m/s. The fundamental frequency is f = v/(2L) = 163/(2 × 0.65) = 125 Hz, more precisely f = 125.2 Hz.
The beat frequency is |f − 128| = 2.8 Hz, consistent with 3 beats/s. Increasing tension increases v and therefore increases f. Since the beat frequency decreases, the wire frequency must be moving toward 128 Hz; it was below 128 Hz.
2. Calculate the tension at which the wire is in unison with the tuning fork. Why is the beat frequency not zero for all tensions near this value?
Read the worked answer
At unison f = 128 Hz. Using f = (1/2L)√(T/μ), T = μ(2Lf)² = (4.0 × 10⁻³)[2 × 0.65 × 128]². The bracket is 166.4 m/s, so T = 4.0 × 10⁻³ × (2.769 × 10⁴) = 110.8 N.
Near unison the beat frequency is proportional to the small frequency difference, not to the tension difference. It passes through zero only at the exact matching tension. The tuning fork itself is specified only to ±0.5 Hz, so the zero-beat tension determines the wire frequency only to within that fork tolerance unless the fork is calibrated.
3. At a tension of 115 N, calculate the beat frequency and state whether the wire is above or below the fork.
Read the worked answer
At 115 N, v = √(115/(4.0 × 10⁻³)) = √(2.875 × 10⁴) = 169.6 m/s, so f = 169.6/1.3 = 130.5 Hz. The beat frequency is 130.5 − 128 = 2.5 Hz.
This tension is above the unison tension (110.8 N), so the wire is now above the fork. The sign of the frequency difference has reversed but the beat frequency is again small; this is why the direction of change, not just the beat count, identifies the original side.