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A stretched wire under tension of 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 mass lg. The vibrating tuning fork in now moved away from the vibrating wire at a constant speed and an observer standing near the sonometer hears one beat per sec. Calculate the speed with which the tuning fork is moved, if the speed of sound in air is 300 m/s. |
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Answer» Solution :As the frequency of a vibrating string is `f=(1)/(2l) sqrt((T)/(mu))` On SUBSTITUTION gives `f=400 Hz`. When the tuning fork is moved away from the observer standing near the sonometer at a constant speed u the apparent frequency of tuning fork will be `f_(R)=f[(V)/(V+u)]` As `f_(R)` is producing beats with f, `f_(R)` is nearly EQUAL to f, i.e., `u lt lt V` so that, `f_(R)=f[1+(u)/(V)]^(-1)=f[1-(u)/(V)]` So beat frequency `DELTAF=f-f_(R)=f|(u)/(V)]` and substituting gives data, `u=V|(Deltaf)/(f)|=300[(1)/(400)]=0.75m//s` |
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