Find the temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion ( alpha) of the material of the tube.

Find the temperature at which the fundamental frequency of an organ pi

1.	Find the temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion ( alpha) of the material of the tube.
Answer» `1//3alpha` `1//2ALPHA` `1//4alpha` `1//5alpha` Solution :Fundamental frequency of an organ pipe, `f=(v)/(2l)` where `l` is length of the organ pipe at temperature `T` Also, `l=l_(0)[1+alpha(T-T_(0))]` VELOCITY of sound, `v=sqrt((gammaRT)/(M))` We have to find the temperatureat which `f(T')=f(T_(0))` for small `(T-T_(0))` `(sqrt((gammaRT)/(M)))/(2l_(0)[1+ alpha(T-T_(0))])=(sqrt ((gammaRT_(0))/(M)))/(2l_(0))rArrsqrt((T)/( T_(0)))=1+alpha(T-T_(0))` `" "[1+((T-T_(0))/(T_(0)))]^(1//2) =1+alpha(T -T_(0))` For small ` (T-T_(0))` we may use binomial APPROXIMATION, `1+(1)/(2)((T-T_(0))/(T_(0 )))= 1+alpha(T-T_(0))` `(1)/(2)((T-T_(0))/(T _(0)))=alpha( T-T_(0)),alpha=(1)/(2T _(0))orT_(0)=(1)/(2alpha)`