Wave Speed on a Stretched String — Question 3

Question 3

An accelerating hanging load does not set tension equal to its weight

A light string is tensioned by a hanging mass $M=2.0\,\mathrm{kg}$ accelerating downward at $a=2.0\,\mathrm{m/s^2}$. Take $g=10\,\mathrm{m/s^2}$ and a test section with $\mu=.010\,\mathrm{kg/m}$. Treat tension as uniform there and neglect string acceleration effects on it.

1. Find tension and transverse wave speed relative to the string.

2. Compare with the static-load prediction and explain what additional information is needed for a laboratory-frame travel speed if the string itself moves.

Question 3 - Solution

An accelerating hanging load does not set tension equal to its weight

Method and mathematical working

Find the tension from Newton's second law for the accelerating mass before applying the wave-speed law. Take downward as positive.$$\begin{aligned}Mg-F&=Ma,\qquad F=M(g-a),\\v&=\sqrt{\frac{M(g-a)}{\mu}}.\\[4pt]F&=2.0(10-2)=16\,\mathrm N,\qquad v=\sqrt{16/0.010}=40\,\mathrm{m/s}.\end{aligned}$$

(a) Calculation and conclusion

For the load, $Mg-F=Ma$, giving $F=M(g-a)=16\,\mathrm{N}$. Thus $v=\sqrt{16/.010}=40\,\mathrm{m/s}$. Static balance would give $20\,\mathrm{N}$ and $44.72\,\mathrm{m/s}$.

(b) Interpretation and limits

The static prediction is too large because some weight produces downward acceleration. A moving string carries the wave relative to the laboratory; its local longitudinal velocity and propagation direction are needed to transform the relative speed. The quoted $40\,\mathrm{m/s}$ is relative to material.

Physical check

For downward acceleration the tension is smaller than $Mg$, and the wave speed must be smaller than the static prediction. The limiting case $a\to g$ removes the supporting tension.