Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy w.r.t. instantaneous displacement.
Two sound notes have wavelengths `(83)/(170)m and (83)/(172)m` in the air. These notes when sounded together produce 8 beats per second. Calculate the velocity of sound in the air and frequencies of the two notes.

Explanation

Law of conservation of energy in S.H.M. :
Let a particle of mass m performing S.H.M. along the path AB about mean position O.
`OB=OA=a`
Point Q is at distance x from mean position O.

Particle performing S.H.M.
At distance x from mean position:
`P.E.=(1)/(2)m omega^(2) x^(2)`
`=(1)/(2)kx^(2)" " [because k=m omega^(2)]`
`K.E. =(1)/(2)mv^(2)`
`=(1)/(2)m omega^(2)(a^(2)-x^(2))`
`=(1)/(2)k(a^(2)-x^(2)) " "[because v=omega sqrt(a^(2)-x^(2))]`
`therefore ` Total energy (T.E.) =K.E. + P.E.
`T.E. =(1)/(2)k(a^(2)-x^(2))+(1)/(2)kx^(2)`
`T.E.=(1)/(2)ka^(2)`
`T.E.=E=(1)/(2)m omega^(2)a^(2) " " `....(i)
At mean position O :
At O, velocity of particle is maximum and displacement is minimum i.e., x = 0.
`therefore P.E. =(1)/(2) kx^(2)=0`
`K.E. =(1)/(2)m omega^(2)(a^(2)-x^(2))`
`=(1)/(2) m omega^(2)( a^(2)-0^(2))`
`therefore K.E. =(1)/(2)m omega^(2)a^(2)`
`=(1)/(2) ka^(2)`
`therefore` Total energy `(T.E.) =KE + P.E. `
`therefore T.E. =(1)/(2)ka^(2)+0=(1)/(2)ka^(2)`
`therefore T.E.=E=(1)/(2)m omega^(2) a^(2) " " `...(ii)
At extreme position A :
At A, velocity of particle is minimum and displacement is maximum i.e., `x=pm a`
`therefore P.E. =(1)/(2)ka^(2)=(1)/(2) m omega^(2)a^(2)`
`K.E. =(1)/(2)m omega^(2)(a^(2)-a^(2))=0`
`therefore `Total energy `(T.E.)=K.E.+P.E.`
`therefore T.E. =0+(1)/(2)ka^(2)`
`therefore T.E. =(1)/(2)ka^(2)`
`therefore T.E. =E=(1)/(2)m omega^(2)a^(2) " " ` ...(iii)
From equation (i), (ii), and (iii), we can say that total energy of S.H.M. at any position is conserved.
Variation of K.E. and P.E. w.r.t. instantaneous displacement is shown below (graphically):

At mean position , total energy is purely kinetic and at extreme position, total energy is purely potential. At other positions, the K.E. and P.E. are interconvertible to each other and their sum is constant which is `(1)/(2)ka^(2)`.
Numerical : We have, Beat frequency = 8
`lambda_(1)=(83)/(170)m and lambda_(2)=(83)/(172)m.`
`lambda_(1) gt lambda_(2)`
`therefore n_(2) gt n_(1)`
`therefore n_(2) -n_(1)=8`
`therefore n_(2)=8+n_(1)`
`n_(1)lambda_(1)=n_(2)lambda_(2)`
`n_(1) xx (83)/(170)=(8+n_(1))(83)/(172)`
`(n_(1))/(170)=(8)/(172)+(n_(1))/(172)`
`172n_(1)=1360+170n_(1)`
`172 n_(1)-170n_(1) =1360`
`2 n_(1)=1360`
`n_(1)=680Hz`
`and n_(2)=(680 +8)Hz=688Hz.`
`therefore ` Frequencies are 680Hz and 688 Hz.
Now, `v=n_(1)lambda_(1)`
`=680 xx (83)/(170)`
`v=332 m//s`