Question 1:
easyA particle is executing a simple harmonic motion. Its maximum acceleration is $ \alpha $ and maximum velocity is $ \beta $. Then, its time period of vibration will be:
(2015 Re)
Maximum acceleration is $ a_{\text{max}} = A\omega^2 = \alpha $ and maximum velocity is $ v_{\text{max}} = A\omega = \beta $. Dividing them, we get $ \omega = \frac{\alpha}{\beta} $. Thus, the time period is $ T = \frac{2\pi}{\omega} = \frac{2\pi \beta}{\alpha} $.
Question 2:
easyA body is executing simple harmonic motion with frequency ‘$n$’, the frequency of its potential energy is:
(2021)
In SHM, the displacement is $x = A\sin(\omega t)$. The potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(\omega t)$.\nSince $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$, the frequency of PE is twice the frequency of displacement, so $2n$.
Question 3:
moderateThe particle executing simple harmonic motion has a kinetic energy $K_0 \cos^2 \omega t$. The maximum values of the potential energy and the total energy are respectively:
(2007)
The maximum kinetic energy is $K_0$.\nIn an ideal SHM without damping, the total energy remains conserved and equals the maximum kinetic energy, which is $K_0$.\nThe maximum potential energy is also equal to the total energy, which is $K_0$.
Question 4:
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 5:
easyThe total energy of particle performing S.H.M. depend on:
(2001)
The total energy in S.H.M is $E = \frac{1}{2} K a^2$. Thus, it depends only on the force constant $K$ and amplitude $a$.
Question 6:
easyA linear harmonic oscillator of force constant $2 \times 10^6 \text{ N/m}$ and amplitude $0.01 \text{ m}$ has a total mechanical energy of $160 \text{ J}$. Its
(1996)
Max K.E. = $\frac{1}{2} k a^2 = \frac{1}{2} \times 2 \times 10^6 \times (0.01)^2 = 100 \text{ J}$. Total Energy = $160 \text{ J}$. Min P.E. = $160 - 100 = 60 \text{ J}$. Max P.E. = Total Energy = $160 \text{ J}$.
Question 7:
easyIn a simple harmonic motion, when the displacement is one-half of the amplitude, what fraction of the total energy is kinetic?
(1995)
$K.E. = \frac{1}{2} k (a^2 - x^2)$. At $x = a/2$, $K.E. = \frac{1}{2} k (a^2 - a^2/4) = \frac{3}{4} (\frac{1}{2} k a^2) = \frac{3}{4} E$.
Question 8:
easyA body executes simple harmonic motion with an amplitude $A$. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy?
(1992)
$P.E. = \frac{1}{4} E \Rightarrow \frac{1}{2} k x^2 = \frac{1}{4} (\frac{1}{2} k A^2) \Rightarrow x^2 = \frac{A^2}{4} \Rightarrow x = \frac{A}{2}$.
Question 9:
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 10:
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} $.