Practice free Oscillations (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
According to the force law for SHM, if the displacement 'x' is positive, the restoring force 'F' will be:
The force law is F(t) = -kx(t). If x is positive, then F = -k(positive value), which results in a negative force. This indicates the force acts in the opposite direction to the displacement, towards the mean position. (NCERT, page 267-268)
The $x$-$t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2\ \text{s}$ is:
Period $T = 8$ s, $\omega = \pi/4$. At $t = 2$ s, $x = -1$ m (minimum). $a = -\omega^2 x = (\pi/4)^2 = \pi^2/16\ \text{m s}^{-2}$.
If $x = 5\sin\left(\pi t + \dfrac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are:
Amplitude 5 m; $T = 2\pi/\pi = 2$ s.
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\dfrac{x}{2}$ times its original time period. Then the value of $x$ is:
$T \propto \sqrt L$, mass-independent. $T_\text{new}/T_\text{old} = \sqrt{1/2} = x/2 \Rightarrow x = \sqrt 2$.
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
As sand leaks, mass $m$ decreases, so $\omega = \sqrt{k/m}$ increases. The adiabatic invariant $E/\omega$ is constant and $E = \tfrac12 kA^2$, giving $A\propto\sqrt{\omega}$, so $A$ also increases.
Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $(A_Q/A_P)$ of the amplitude $A_Q$ of mass Q to the amplitude $A_P$ of mass P is:
$v_{max} = A\omega = A\sqrt{k/m}$. Equal masses and equal $v_{max}$: $A_P\sqrt{k_1} = A_Q\sqrt{k_2}\Rightarrow\dfrac{A_Q}{A_P} = \sqrt{\dfrac{k_1}{k_2}}$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.