A straight conducting bar of mass m and length λ 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)
Text Solution
Verified by ExpertsA
As capacitor discharge in time Δ t, a current will flow due to which magnetic force will act on bar for small time.
Impulse of force
= F Δ t
= i λ B Δ t
= 
λ B Δ t
= B λ Δ q
Δ q is the total charge flowing through capacitor in time Δ t.
Change in momentum = mv
mv = B λ Δ q
mv = B λ CU ... (1)
[ Δ q was the charge in C i.e. CU]
two spring are in parallel
equivalent stiffness constant = 2k
Energy conservation
mv 2 = 
where A is amplitude
A = 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems