Question

A violin string tuned to 485.0 Hz is found to exhibit 2.6 beats per second with a guitar string. The guitar string is then tightened, and the beat frequency is found to increase. What was the original frequency of the guitar string? (Enter your answer to one decimal place.)

Answer 1

Step-by-step explanation:

It is given that,

Frequency of violin string tuned to,
`f_1=485\ Hz`

Beat frequency,
`\beta=2.6\ beats/s`

Number of beats per second is given by :

`|f_1-f_2|=2.6`

`|485-f_2|=2.6`

`f_2=485\pm2.6`

Either
`f_2=482.4\ Hz` or
`f_2=487.6\ Hz`

When the guitar string is tightened, and the beat frequency is found to increase, then the original frequency will decrease.

So, the original frequency of the guitar string is 482.4 Hz. Hence, this is the required solution.