In the figure below, a string, tied to a sinusoidal oscillator at P and running
ID: 2012557 • Letter: I
Question
In the figure below, a string, tied to a sinusoidal oscillator at P and running over a support at Q, is stretched by a block of mass m. The separation L between P and Q is 1.80 m, and the frequency f of the oscillator is fixed at 120 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging block is 286.1 g or 643.7 g, but not for any intermediate mass. What is the linear density of the string?
Explanation / Answer
The block is in equilibrium. Thus thetension in the string is
0 = F – mg F = mg
To produce a standing wave in the
string the mass of the block must satisfy
f = nv/(2L) = (n/2L)(F/) = (n/2L)(mg/) m = 4L^2f^2/(n^2g)
Since mass m is inversely proportional to the harmonic number n, if you increase the mass of the
block, the harmonic number decreases. Therefore, if m = 643.7 g creates a standing wave with
harmonic number n, then m’ = 286.1 g creates a standing wave with harmonic number n + 1. Taking
the ratio of two masses, the harmonic number n is
m/m’ = (n + 1)^2/n^2
n = 1/((m/m’) – 1) = 2
Using the expression for the mass m obtained in part a), we get
m = 4L^2f^2/(n^2g) = mn^2g/(4L^2f^2) = 1.35 * 10-4 kg /m = 0.135 g/m = 0.135 g/m