First slide

Beats

Question

Two wires are in unison. If the tension in one of the wires is increased by 2%, 5 beats are produced per second. The initial frequency of each wire is

Moderate

A

200 Hz
B

400 Hz
C

500 Hz
D

1000 Hz

Solution

$n \propto \sqrt{T} \Rightarrow \frac{Δn}{n} = \frac{ΔT}{2 T}$
If tension increases by 2%, then frequency must increases by 1%.
If initial frequency n₁ = n then final frequency n₂ – n₁ = 5
$\Rightarrow \frac{101}{100} n - n = 5 \Rightarrow n = 500 Hz .$
Short trick : If you can remember then apply following formula to solve such type of problems.
Initial frequency of each wire (n)
$= \frac{(Number of beats heard per sec) \times 200}{(per centage change in tension of the wire)}$
Here $n = \frac{5 \times 200}{2} = 500 Hz$

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