2 Marks Questions

Question

Show that for a particle performing linear simple harmonic motion, the average kinetic energy of any period of oscillation is equal to the average potential energy of the same period.

Vidyadip · Accepted Answer

Suppose a particle of mass performs simple harmonic motion for time period T. At the instant when time is measured from the mean position, the displacement of the particle is y=a t (1) We know that v & = Velocity of particle & =d y d t =d d t (a t) & =a t (2) , E_k & =1 2 m v^2=1 2 m(a t)^2 & =1 2 m a^2 ^2 ^2 t P.E., E_p & =1 2 ky^2=1 2 k(a t)^2 _p & =1 2 ka^2 ^2 t & =1 2 m ^2 a^2 ^2 t (E_k )_ a v & =Average kinetic energy of one cycle & =1 T _0^ T E_k d t & =1 T _0^ T (1 2 m a^2 ^2 ^2 t ) d t & =1 2 T m a^2 ^2 _0^ T ^2 t d t & =1 2 T m a^2 ^2 _0^ T (1+ 2 t 2 ) d t & =1 4 T m a^2 ^2 [ _0^ T 1 d t+ _0^ T 2 t d t ] & =1 4 T m a^2 ^2 [(T-0)+ ( 2 t 2 )_0^ T ] & =ma^2 ^2 4 T [T+1 2 ( 4 T T- 0 ) ] & =ma^2 ^2 4 T [T+1 2 (0-0) ] n & =0, n=0,1,2 (E_i )_ a v & =1 4 ma^2 ^2 (3) Average potential energy one cycle l (E_ p )_ a v =1 T _0^ T E_ p d t given by (E_ p )_ av =1 T _0^ T 1 2 m ^2 a^2 ^2 t d t =m ^2 a^2 2 T _0^ T ^2 t d t =m ^2 a^2 2 T _0^T (1- 2 t 2 ) d t 2 =1-2 ^2 =1 4 T m ^2 a^2 [ _0^ T 1 . d t- _0^ T 2 d t ] =1 4 T m ^2 a^2 [(T-0)- ( 2 t 2 )_0^ T ] (E_ p )_ a v =1 4 m a^2 ^2 (4) Thus, from equation (3) and (4) we see that the average K.E. over one oscillation time period is and during the same period the average P.E. is equiralent to.

Answer

Need a full question paper?

Explore more

Similar questions

PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions