question_answer

A stretched string of length 1 m and mass \[\text{5}\times \text{1}{{0}^{-4}}\text{ kg}\] fixed at both ends is under tension of 20 N. it is plucked at a point situated at 25 cm from one end. The frequency of vibration of the string will be

A)  125 Hz                                 

B)  50 Hz

C)  200 Hz                                 

D)   100 Hz

Correct Answer: C

Solution :

                 The frequency (n) of vibration of string is given by                 \[n=\frac{p}{2\,l}\sqrt{\frac{T}{m}}\] where T is tension, and m the mass per unit length.                                 Given,   \[p=2,\,\,\,m=5\times {{10}^{-4}}kg\],                 T = 20N,                \[l=1\,m\] \[\therefore \]  \[n=\frac{2}{2\times 1}\sqrt{\frac{20}{5\times {{10}^{-4}}}}=200\,Hz\]