Question

A trolley of mass m is connected to two identical springs, each of force constant k, as shown in figure. The trolley is displaced from its equilibrium position by a distance x and released. The trolley executes simple harmonic motion of period T. After some time it comes to rest due to friction. The total energy dissipated as heat is (assume the damping force to be weak)

• A.
• B.
• C.
• D.

Answer 1

The correct option are

B

C $\frac{2{\pi }^{2}m{x}^2}{T^2}$
In general, the motion of a damped oscillator is not simple harmonic. If the damping forces are weak, the motion is very nearly simple harmonic and all formulae of SHM apply.

The amplitude A = x.

The time period T is $T=2\pi \sqrt{\frac{m}{2k}}$ ..........(i)

If the trolley eventually comes to rest, the entire energy of oscillation is dissipated as heat due to friction. Hence, the total energy dissipated as heat is
$E=\frac{1}{2}m{A}^2{\omega }^2=\frac{1}{2}m{A}^2{\left(\frac{2\pi }{T}\right)}^2=\frac{2{\pi }^{2}m{A}^2}{T^2}$

$E=\frac{1}{2}m{A}^2{\omega }^2=\frac{1}{2}m{x}^2{\left(\frac{2\pi }{T}\right)}^2=\frac{2{\pi }^{2}m{x}^2}{T^2}$

(ii) Which is choice (C).
Using (i) in (ii) we get $E=k{x}^2$
Hence choice (B) is also correct.