Stationary waves on a string

What you are trying to do

Find how the first-harmonic frequency of a stationary wave on a stretched string depends on the string's length, the tension in it, and its mass per unit length.

Apparatus

• Signal generator driving a vibration generator, clamped at one end of the bench
• String or wire over a pulley at the far end, with a hanger and slotted masses providing the tension
• A movable bridge to set the vibrating length, and a metre rule
• A top-pan balance and a long sample of the same string, for the mass per unit length

Variables

• Independent: the vibrating length L (then, in a second series, the tension T)
• Dependent: the first-harmonic frequency
• Control: whichever of T and L is not being varied, and the string itself, so the mass per unit length stays fixed

Method

1. Measure the mass per unit length first: weigh a long measured sample of the string and divide. A long sample keeps the percentage uncertainty down.
2. Hang a known mass over the pulley; the tension is its weight.

1. Set the bridge for the chosen length, then sweep the signal generator slowly until the string vibrates in one large loop. Approach the resonance from below and from above and take the middle of the range that looks maximal: judging the peak is the main skill of the experiment.
2. Record the frequency, then repeat for a series of lengths at fixed tension, and afterwards for a series of tensions at fixed length.

Analysis

1. The first harmonic satisfies f = (1/2L) × the square root of T/μ, so at fixed tension f is proportional to 1/L.
2. Plot f against 1/L: a straight line through the origin whose gradient is half the wave speed. From the gradient, recover the speed, and from v² = T/μ recover the mass per unit length to compare with your weighed value.
3. For the tension series, plot f against the square root of T: straight through the origin again if the model holds.

A worked set of readings

String of mass per unit length 4.8 × 10⁻⁴ kg m⁻¹ under a 200 g load (tension 1.96 N):

The gradient of f against 1/L is 32.0 Hz m, so the wave speed is v = 2 × 32.0 = 64 m s⁻¹. Then μ = T/v² = 1.96/64² = 4.8 × 10⁻⁴ kg m⁻¹, matching the weighed value: the model closes on itself.

Where the uncertainty comes from

• Judging the resonance: The amplitude peak is broad, so bracket it: note the frequency where the amplitude clearly falls off on each side and take the midpoint and half-range.
• The vibrating length: Measured between the bridge and the vibration generator; neither end is a perfect node, so keep L large to make the end effects proportionally small.
• Mass per unit length: Weigh a metre or more, not a short offcut; the balance reads to 0.01 g, and a short sample wastes that.
• Tension: Assumes a frictionless pulley and a stationary hanging mass; check the mass hangs freely and is not oscillating.

What earns the marks

Safety