Question 11:
moderateThe potential energy of a simple harmonic oscillator when the particle is half way to its end point is:
(2003)
Total energy $E = \frac{1}{2}kA^2$. Halfway to the endpoint means $x = A/2$.\nPotential energy $$U = \frac{1}{2}kx^2 = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2)$$.\nTherefore, $U = E/4$.
Question 12:
moderateThe bob of simple pendulum having length is displaced from mean position to an angular position $ \theta $ with respect to vertical. If it is released, then velocity of bob at lowest position:
(2000)
Change in potential energy equals kinetic energy at the lowest point. $ mgl(1-\cos\theta) = \frac{1}{2}mv^2 $. Solving for velocity gives $ v = \sqrt{2gl(1-\cos\theta)} $. (Note: length parameter $ l $ is implied in option a despite typo).
Question 13:
moderateTwo sphrical bob of masses $ M_A $ and $ M_B $ are hung vertically from two strings of length $ \ell_A $ and $ \ell_B $ respectively. They are executing SHM with frequency relation $ f_A = 2f_B $, Then:
(2000)
Frequency of a simple pendulum is $ f = \frac{1}{2\pi}\sqrt{\frac{g}{\ell}} $, which is independent of mass. Given $ f_A = 2f_B $, we have $ \frac{1}{\sqrt{\ell_A}} = \frac{2}{\sqrt{\ell_B}} $. Squaring both sides yields $ \ell_A = \frac{\ell_B}{4} $.
Question 14:
moderateA loaded vertical spring executes S.H.M. with a time period of 4 sec. The difference between the kinetic energy and potential energy of this system varies with a period of:
(1994)
In SHM, kinetic and potential energies (and their difference) oscillate with twice the frequency of the displacement. Therefore, their period is half of the displacement period, $ T' = T/2 = 4/2 = 2 $ sec.