Question

A straight conducting bar of mass m and length l is suspended horizontally with two non-conducting springs of stiffness k as shown in figure. The capacitor is initially charged to the potential difference U. At time t = 0, the switch S is closed and the capacitor discharges. The bar starts oscillating in vertical plane.

Find the amplitude of these oscillation : (Assume the time of discharge of capacitor is much smaller than the period T of the mechanical oscillation of the bar)

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Answer 2

As capacitor discharge in time Dt, a current will flow due to which magnetic force will act on bar for small time.

Impulse of force

= F Dt

= ilBDt

=lBDt

= BlDq

Dq is the total charge flowing through capacitor in time Dt.

Change in momentum = mv

mv = BlDq

mv = BlCU ... (1)

[Dq was the charge in C i.e. CU]

two spring are in parallel

$\therefore$equivalent stiffness constant = 2k

Energy conservation

$\frac { 1 } { 2 }$mv² = $\frac { 1 } { 2 } \times 2 \mathrm { kA } ^ { 2 }$

where A is amplitude

A =