Question

Two strings have different lengths and linear densities, as the drawing shows. They are joined together and stretched so that the tension in each string is
. The free ends of the joined string are fixed in place. Find the lowest frequency that permits standing waves in both strings with a node at the junction. The standing wave pattern in each string may have a different number of lo

Answer 1

Let the two string segments have lengths L1, L2 and linear mass densities μ1, μ2, and let the common tension be T. The wave speeds are
v1 = sqrt(T/μ1), v2 = sqrt(T/μ2).

A node at the junction and fixed ends at the two free ends imply that each segment must contain an integer number of half‑wave‑lengths:
L1 = n1(λ1/2), L2 = n2(λ2/2), with n1,n2 ∈ {1,2,3,...}.
Using λi = vi/f gives
f = n1 v1/(2 L1) = n2 v2/(2 L2).

Thus the allowed common frequencies are those for which
n1 v1/L1 = n2 v2/L2.
Or equivalently
n1/n2 = (v2 L1)/(v1 L2) = (L1/L2) sqrt(μ1/μ2).

The lowest nonzero frequency with a node at the junction is obtained by choosing the smallest positive integers n1, n2 that satisfy the above equality, and then
f_min = n1 v1/(2 L1) = n2 v2/(2 L2).

Remark: if the ratio (L1/L2) sqrt(μ1/μ2) is irrational there are no nonzero integers n1,n2 that satisfy the equality exactly, so there is no exact nonzero common standing‑wave frequency with a node exactly at the junction.