A 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.
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $ T = 2\pi \sqrt{(l/g’)} $, where $ g’ $ is equal to: (1991)
For a trolley accelerating horizontally, the effective gravity is the vector sum of standard gravity $ g $ (downwards) and the pseudo acceleration $ a $ (backwards). Hence, $ g' = \sqrt{g^2 + a^2} $.
The angular velocity and the amplitude of a simple pendulum is $ \omega $ and a respectively. At a displacement $ x $ from the mean position if its kinetic energy is $ T $ and potential energy is $ V $, then the ratio of $ T $ to $ V $ is: (1991)
Kinetic energy is $ T = \frac{1}{2}m\omega^2(a^2 - x^2) $ and potential energy is $ V = \frac{1}{2}m\omega^2x^2 $. Taking the ratio gives $ \frac{T}{V} = \frac{a^2 - x^2}{x^2} $.
A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $ k_1 $ and $ k_2 $. The time period of the suspended mass will be: (1990)
For two springs connected in series, the equivalent spring constant is $ k_{eq} = \frac{k_1k_2}{k_1 + k_2} $. Substituting this into the time period formula $ T = 2\pi \sqrt{\frac{m}{k_{eq}}} $ yields $ T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1k_2}} $.
The 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).
Two 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} $.
A spring elongated by length L when a mass M is suspended to it. Now a tiny mass m is attached and then released, its time period of oscillation is: (1999)
The spring constant is $ k = \frac{Mg}{L} $. When total mass becomes $ (M+m) $, the time period is $ T = 2\pi \sqrt{\frac{M+m}{k}} = 2\pi \sqrt{\frac{(M+m)L}{Mg}} $.
Frequency of simple pendulum in a free falling lift is: (1999)
In a free-falling lift, the effective acceleration due to gravity is zero ($ g_{eff} = g - g = 0 $). Since frequency $ f = \frac{1}{2\pi}\sqrt{\frac{g_{eff}}{l}} $, the frequency becomes zero.
Two pendulums suspended from same point having length 2 m and 0.5 m. If they displaced slightly and released then they will be in same phase, when small pendulum will have completed: (1998)
Time period $ T \propto \sqrt{l} $. $ \frac{T_1}{T_2} = \sqrt{\frac{2}{0.5}} = 2 $, so $ T_1 = 2T_2 $. They return to the same phase when the longer pendulum completes 1 oscillation and the shorter one completes 2 oscillations.
If the frequency of a spring is n after suspending mass M, now 4M mass is suspended from spring then the frequency will be: (1998)
Frequency of a spring-mass system is $ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \implies f \propto \frac{1}{\sqrt{m}} $. If mass is increased to 4M, the new frequency is $ f' = \frac{f}{\sqrt{4}} = \frac{n}{2} $.