Question 41:
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 42:
moderateThe equations of two waves given as $x = a\cos(\omega t + \delta)$ and $y = a\cos(\omega t + \alpha)$, Where $\delta = \alpha + \pi/2$, then resultant wave represent:
(2000)
Substitute $\delta$: $x = a\cos(\omega t + \alpha + \pi/2) = -a\sin(\omega t + \alpha)$.\nCombining with $y = a\cos(\omega t + \alpha)$ yields $x^2 + y^2 = a^2$, a circle.\nThe direction of motion traces anti-clockwise as time progresses.
Question 43:
moderateA simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $X = A$ to $X = A/2$ is:
(1992)
Using equation for SHM starting from extreme position: $x = A\cos(\omega t)$.\nSubstitute $x = A/2$: $A/2 = A\cos(2\pi t/T) \implies \cos(2\pi t/T) = 1/2$.\n$2\pi t/T = \pi/3 \implies t = T/6$.
Question 44:
moderateTwo pendulums of length $121 \text{ cm}$ and $100 \text{ cm}$ start vibrating in phase. At some instant, the two are at their means position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the means position is :
(2022)
$T \propto \sqrt{l}$. So $T_1/T_2 = \sqrt{121/100} = 11/10$. This gives $10 T_1 = 11 T_2$. The shorter pendulum ($T_2$) completes $11$ vibrations.
Question 45:
moderateA spring is stretched by $5 \text{ cm}$ by a force $10 \text{ N}$. The time period of the oscillations when a mass of $2 \text{ kg}$ is suspended by it is:
(2021)
$k = F/x = 10 / 0.05 = 200 \text{ N/m}$. Time period $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{2/200} = \frac{2\pi}{10} = 0.628 \text{ s}$.
Question 46:
moderateA pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is $20 \text{ m/s}^2$ at a distance of $5 \text{ m}$ from the mean position. The time period of oscillation is:
(2018)
Acceleration $a = \omega^2 x \Rightarrow 20 = \omega^2 (5) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \text{ rad/s}$. Time period $T = \frac{2\pi}{\omega} = \pi \text{ s}$.
Question 47:
moderateA spring of force constant $k$ is cut into lengths of ratio $1 : 2 : 3$. They are connected in series and the new force constant is $K’$. Then they are connected in parallel and force constant is $K”$. Then $K’ : K”$ is:
(2017-Delhi)
$k \propto 1/L$. The parts have stiffness $6k, 3k, 2k$. In series, $K' = k$. In parallel, $K'' = 6k + 3k + 2k = 11k$. The ratio is $1:11$.
Question 48:
moderateA body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3 \text{ s}$. When the mass $m$ is increased by $1 \text{ kg}$, the time period of oscillations becomes $5 \text{ s}$. The value of $m$ in kg is:
(2016 – II)
$T \propto \sqrt{m} \Rightarrow \frac{3}{5} = \sqrt{\frac{m}{m+1}} \Rightarrow \frac{9}{25} = \frac{m}{m+1} \Rightarrow 25m = 9m + 9 \Rightarrow m = \frac{9}{16} \text{ kg}$.
Question 49:
moderateThe 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} $.
Question 50:
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).