Practice portal › Waves › Standing Waves in Strings

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is

Asked in NEET 2015 · Nodes, antinodes and harmonics

Answer: (2) 105 Hz

Step-by-step solution

Given: consecutive resonances at 315 Hz and 420 Hz on a string fixed at both ends.

Idea: such a string resonates at uₙ=(nv)/(2L) for every whole n, so consecutive resonances differ by v/(2L) — which is the fundamental itself.

Δ u= uₙ₊₁- uₙ=v/(2L)= u₁.

Δ u=420-315.

u₁=105 Hz.

(Checking: 315=3×105 and 420=4×105, so these are the 3rd and 4th harmonics, with nothing between them.)

Why the other options are wrong

More Standing Waves in Strings questionsAll Standing Waves in Strings questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer