At `16^@C` , two open organ pipes , when sounded together, produce 34 beats in 2 seconds. How many beats per second will be produced, if the temperature rises to `51^@C`? Neglect the increase in the length of the pipes.

Answer

18

Explanation

Number of beats produced per second at `16^@C = 34//2=17`
Let `L_1, L_2 =` Lengths of the two organ pipes,
`v_1,v_2` = Lowest frequencies emitted by the pipes at `16^@C`
`v._(1)v._(2)` = Lowest frequencies emitted by the pipes at `51^@C`
`v_(16),v_(51)` = velocities of sound at `16 ^@C and 51^@C` respectively.
`:.v_1=(v_(16))/(2L_1) and v_2=(v_(16))/(2L_2) `
Beat frequency at `16^@C`
`v_1 - v_2 = 17`
or `(v_(16))/(2L_1)-(v_(16))/(2L_2)=17`
`v_(16)[1/(2L_1)-1/(2L_2)]=17" "(i)`
Let the number of beats per second at `51^@C=b`
Then `v._(1) - v._(2) =b`
or `v_(51)[1/(2L_1)-1/(2L_2)]=b" "...(ii)`
Dividing equation (ii) by (i), we get
`(V_(51))/(v_(16))=b/17` or `sqrt((51+273)/(16+273))=b/17" " [ :. v_t prop sqrt(t+273)]`
`b = 17 xxsqrt((324)/(289))=17 xx(18)/(17)=18s^(-1)`