Question

A stretched wire emits a fundamental note of 256 Hz. Keeping the stretching force constant and reducing the length of wire by 10 cm, the frequency becomes 320 Hz. Calculate the original length of the wire.

Answer 1

frequency of fundamental note is given by
$\bold{\nu=\frac{v}{2l}\sqrt{\frac{T}{\mu}}}$
Here, ν is the frequency of fundamental note , v is speed of wave , l is the length of string , T is tension in the string and μ is mass density { mass per unit length}

in first case ,
$\bold{256Hz=\frac{v}{2l}\sqrt{\frac{T}{\mu}}}$---(1)

in second case,
$\bold{320Hz=\frac{v}{2(l-10)}\sqrt{\frac{T}{\mu}}}$----(2)

From equations (1) and (2)
256/320 = 2(l-10)/2l
⇒ 4/5 = ( l - 10)/l
⇒ 4l = 5( l - 10)
⇒ 4l = 5l - 50 ⇒ l = 50 cm

Hence, length of string is 50 cm