A sonometer wire under tension 64 N , vibrating in its fundamental mode, is in resonance with a vibr

Question

A sonometer wire under tension 64 N , vibrating in its fundamental mode, is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of 10 cm and a mass of 1 g . The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near the sonometer hears one beat per second. Calculate the speed with which the tuning fork is moved if the speed of sound in air is $300 \mathrm{~ms}^{-1}$.\

Professor PC Thomas Classes · Answer

For fundamental frequency of wire,\ $f=\frac{v}{\lambda}=\frac{v}{2 l} \quad 	herefore f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$\ $\mu=\frac{	ext { Mass }}{	ext { length }}=\frac{1 	imes 10^{-3}}{10 	imes 10^{-2}}=0.01 \frac{\mathrm{~kg}}{\mathrm{~m}}$\ $	herefore f=\frac{1}{2 	imes 0.1} \sqrt{\frac{64}{0.01}}$ or $f=400 \mathrm{~Hz}$\ Beat frequency $f_{b}=1$\ $	herefore \quad$ Apparent frequency of fork $$ =400-1=399 \mathrm{~Hz} $$ According to Doppler's effect\ $f^{\prime}=f \frac{v}{\left(v+v_{S}ight)}$\ where $v_{S}=$ speed of tuning fork.\ $	herefore 399=\frac{400 	imes 300}{300+v_{S}}$\ or $399 v_{S}=400 	imes 300-399 	imes 300$\ or $399 v_{S}=300$\ or $v_{S}=\frac{300}{399}$ or $v_{S}=0.75 \mathrm{~m} / \mathrm{s}$\ $	herefore \quad$ Speed of tuning fork $=0.75 \mathrm{~m} / \mathrm{s}$\