A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be

Question

A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be (A) 0.25 m (B) 0.5 m (C) 1 m (D) 2 m

Tardigrade Admin · Accepted Answer

f=(1/2 l)[(T/μ)](1/2) when f is halved, the length is doubled ∴ Length is 1 m.

Q. A stretched string resonates with tuning fork frequency $512 Hz$ when length of the string is $0.5 m$ . The length of the string required to vibrate resonantly with a tuning fork of frequency $256 Hz$ would be

0.25 m

0.5 m

1 m

2 m

$f = \frac{1}{2 l} [\frac{T}{μ}]^{\frac{1}{2}}$ when $f$ is halved, the length is doubled
$∴$ Length is $1 m$ .

Q. A stretched string resonates with tuning fork frequency 512Hz when length of the string is 0.5m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256Hz would be

0.25 m

0.5 m

1 m

2 m

f=2l1​[μT​]21​ when f is halved, the length is doubled ∴ Length is 1m.

Q. A stretched string resonates with tuning fork frequency $512 Hz$ when length of the string is $0.5 m$ . The length of the string required to vibrate resonantly with a tuning fork of frequency $256 Hz$ would be

$f = \frac{1}{2 l} [\frac{T}{μ}]^{\frac{1}{2}}$ when $f$ is halved, the length is doubled
$∴$ Length is $1 m$ .