Question

A wire under tension vibrates with fundamental
frequency of 256 Hz. What should be the fundamental
frequency if the wire half as long as, twice as thick, and
under one-fourth of the tension?​

Answer 1

Answer:

128 Hz

Explanation:

Fundamental frequency of a wire of Length "L", mass per unit density "m", and under a tension "T" is given as

• f = 1/(2L) × √(T/m)

After using simple relations of mass, volume, density and area we get the following formula (see image for derivation)

• f = 1/(2Lr) × √[T / (πρ)]

Where

• r = Radius
• ρ = Density of wire

In first case

256 Hz = 1/(2Lr) × √[T / (πρ)]

In second case

• Length halved (L/2)
• Thickness doubled which means radius is doubled (2r)
• Tension is reduced to ¼th (T/4)

f' = 1/[2 × (L/2) × 2r] × √[(T/4) / (πρ)]

f' = 1/(2Lr) × √[(T/4) / (πρ)]

Divide above two equations

256 Hz / f' = {1/(2Lr) × √[T / (πρ)]} / {1/(2Lr) × √[(T/4) / (πρ)]}

256 Hz / f' = √[1 / (1/4)]

256 Hz / f' = √4

256 Hz / f' = 2

f' = 256 Hz / 2

f' = 128 Hz