A trolley of mass M is connected to two springs of force constant `K_(1)` and `K_(2)` as shown in the figure When the trolley is displacement from its equilibrium postion by a distance a and released , it executes S.H.M of frequency f. If it comes to rest after somtimes due to friction, the total energy dissipated as heat is (Dampling forces are weak)

Options

A. `pi^(2)ma^(2)f^(2)`
B. `2pi^(2)ma^(2)f^(2)`
C. `2a^(2)(K_(1)+K_(2))`
D. `1/2a^(2)(K_(1)+K_(2))`

Answer

B, D

Explanation

The damping forces are weak. So, all formula of S.H.M. apply.
Amplitude = a
Time period
`T = 2pi sqrt((m)/(K_(1) + k_(2)))`
`therefore` Frequency, f = `(1)/(2pi) sqrt((k_(1) + K(_(2))/(m))`
The trolley finally comes to rest, So, the entire energy of oscillation is dissipated as heat due to friction and is given by
`E = (1)/(2)ma^(2)omega^(2) = (1)/(2)ma^(2)(2pif)^(2)`
` = 2pi^(2)ma^(2)f^(2)`
Which is choice (b)
`E = (1)/(2) ma^(2) xx 4pi^(2) xx (1)/(4pi^(2)) ((K_(1) + K_(2))/(m))`
`E = (a^(2)(K_(1) + K_(2))/(2))`
Which is choice (d)