Embibe Experts Solutions for Chapter: Simple Harmonic Motion, Exercise 8: Exercise (Analytical Questions)

The time period of a particle is $8 \mathrm{s}$. At $t=0$, it is at the mean position. The ratio of distance covered by the particle in $1^{st}$ second and $2^{nd}$ second will be:

A man of mass $60 \mathrm{kg}$ standing on a platform executing $\mathrm{S}\text{.}\mathrm{H}\text{.}\mathrm{M}\text{.}$ in the vertical plane. The displacement from the mean position varies as $y=0.5 \sin \left(2\pi ft\right).$ The value of $f,$ for which the man will feel weightlessness at the highest point is ($y$ is in metres)

A point mass oscillates along the $x$-axis according to the law, $x=x_0\cos \left(\mathrm{\omega }t–\frac{\mathrm{\pi }}{4}\right)$. If the acceleration of the particle is written as, $a=A\cos (\omega t+\mathrm{\delta })$, then,

The total energy of a harmonic oscillator of mass $2 \mathrm{kg}$ is $9 \mathrm{joules}$. If its potential energy at mean position is $5 \mathrm{joules}$, its $K.E.$. at the mean position will be:

A horizontal spring is connected to a mass $M$. It executes simple harmonic motion. When the mass $M$ passes through its mean position, an object of mass $m$ is put on it and the two move together. The ratio of frequencies before and after will be:

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$ is given by

A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to the relation $y=k{t}^2,\left(k=1 \mathrm{m} {\mathrm{s}}^{-2}\right)$ where $y$ is the vertical displacement. The time period now becomes $T_2$. The ratio of $\frac{T_1^2}{T_2^2}$ is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

Two waves $Y_1=a\sin \omega t \text{and} Y_2=a\sin \left(\omega t+\delta \right)$ are producing interference. Then, the resultant intensity is,