A wire of length 80 cm has a frequency of 256 Hz. Calculate the length of a similar wire under similar tension, which will have frequency 1024 Hz.
Topic: Sound and vibrations
Answer
Given: l₁ = 80 cm, f₁ = 256 Hz, f₂ = 1024 Hz
The wires are similar and under the same tension.
For a stretched string, the frequency is inversely proportional to the length: f ∝ 1 ÷ l, so f₁l₁ = f₂l₂
256 × 80 = 1024 × l₂ 20 480 = 1024 × l₂ l₂ = 20 480 ÷ 1024 l₂ = 20 cm
Length of the second wire = 20 cm
The law is an INVERSE proportion: shorten the wire and the note goes up. The frequency here rises by a factor of 4 (256 → 1024), so the length must fall by a factor of 4, from 80 cm to 20 cm. You can almost write the answer down without the algebra.
The mistake is treating it as a direct proportion and getting 320 cm. Ask yourself which way it should go before you divide: a longer string gives a LOWER note, like the thick long strings on a guitar.
"Similar wire under similar tension" is the phrase that lets you use f ∝ 1/l on its own; if the tension or the thickness changed, those would enter the formula too.