Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Oscillations, Exercise 2: Critical Thinking

A particle is performing simple harmonic motion with amplitude of $4 \mathrm{cm}$ and time period of $12 \mathrm{s}$. The ratio between time taken by it in going from its mean position to $2 \mathrm{cm}$ and from $2 \mathrm{cm}$ to extreme position is
 

A particle executes simple harmonic motion (amplitude$=A$) between $x=-A \text{and } x=+A$. The time taken for it to go from $0 \mathrm{to} \frac{A}{2}$ is $T_1$ and to go from $\frac{A}{2}\text{ to} A$ is $T_2$.Then
 

A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $A$ metre. The shortest time it takes to reach a point $\frac{A}{\sqrt{2}}\mathrm{m}$ from its mean position in seconds is
 

Assertion: In $\mathrm{S}.\mathrm{H}.\mathrm{M}.$, the velocity and displacement of the particle are in the same phase.

Reason: Velocity is the ratio of displacement to the time taken.

Assertion: If the amplitude of the simple pendulum increases, its time period increases.

Reason: The simple pendulum is to traverse more distance in each vibration when its amplitude is large.

When a particle performs $\mathrm{S}.\mathrm{H}.\mathrm{M}.$, its kinetic energy varies periodically. If the frequency of the particle is $10 \mathrm{Hz}$, then the kinetic energy of the particle will vary with a frequency equal to

A simple pendulum of length $L$ is hanging from rigid support on the ceiling of a stationary train. If the train moves forward with an acceleration $a$, then the time period of the pendulum will be

For a particle executing $\mathrm{S}.\mathrm{H}.\mathrm{M}.$, the displacement $x$ is given by $x=A\cos \left(\omega t\right)$. Identify the graph which represents the variation of potential energy$\left(P.E.\right)$ as a function of time $t$ and displacement $x$.