Question 141:
easyA particle executes S.H.M. along x-axis. The force acting on it is given by:
(1994, 88)
For simple harmonic motion, the restoring force must be proportional to the negative of the displacement.\nThe equation $F = -Akx$ is the only one that satisfies the condition $F \propto -x$.
Question 142:
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 143:
easyIf a simple harmonic oscillator has got a displacement of $0.02 \text{ m}$ and acceleration equal to $2 \text{ m/s}^2$ at any time, the angular frequency of the oscillator is equal to:
(1992)
Magnitude of acceleration in SHM is $|a| = \omega^2|x|$.\nSubstitute the values: $2 = \omega^2 \times 0.02$.\n$\omega^2 = 100 \implies \omega = 10 \text{ rad/s}$.
Question 144:
moderateThe composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along:
(1990)
Let $x = A\sin(\omega t)$ and $y = B\sin(\omega t + \pi) = -B\sin(\omega t)$.\nThen $y/x = -B/A \implies y = -(B/A)x$.\nThis represents the equation of a straight line.
Question 145:
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 146:
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 147:
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 148:
easyThe circular motion of a particle with constant speed is:
(2005)
Uniform circular motion repeats itself after fixed intervals of time, making it periodic.\nHowever, the motion itself is not along a straight line towards a mean position, so it is not simple harmonic.
Question 149:
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 150:
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 $T = \frac{2\pi}{\omega} = \pi \text{ s}$.