Question

When tuning his guitar, a music student notices that the string is not vibrating at the desired frequency of 120 Hz, so he tightens the string by 15.5% to get this frequency. At what frequency was the string vibrating when he first started the tuning process

Answer 1

Answer:

Explanation:

We know that

frequency ∝ √ Tension

So

Δn / n = 1/2 (ΔT / T)

Δn / n = .155 / 2 ( GIVEN )

= .0775

Δn  = .0775 n

Let the original frequency be n

n +Δn = 120

n + .0775 n = 120

n = 111.36