Embibe Experts Solutions for Chapter: Oscillations, Exercise 1: JEE Main - 22 July 2021 Shift 1

The general displacement of a simple harmonic oscillator is $x=A \sin \omega t$. Let $T$ be its time period. The slope of its potential energy ($U$) – time ($t$) curve will be maximum when $t=\frac{T}{\beta }$. The value of $\beta$ is ______.

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \mathrm{kg}$, the angular frequency is ${\omega }_1$. When the mass block is $2 \mathrm{kg}$ the angular frequency is ${\omega }_2$. The ratio $\frac{{\omega }_2}{{\omega }_1}$ is :

The velocity of a particle executing SHM varies with displacement ($x$) as $4v^2=50–x^2$. The time period of oscillations is $\frac{x}{7} \mathrm{s}$. The value of $x$ is ______.

[Take $\pi =\frac{22}{7}$]

The maximum potential energy of a block executing simple harmonic motion is $25 \mathrm{J}$. A is amplitude of oscillation. At $\frac{A}{2}$, the kinetic energy of the block is

In the figure given below, a block of mass $M=490 \mathrm{g}$ placed on a frictionless table is connected with two springs having same spring constant ($K=2 \mathrm{N} {\mathrm{m}}^{-1}$). If the block is horizontally displaced through $X$ $\mathrm{m}$ then the number of complete oscillations it will make in $14\pi$ seconds will be ______.

The amplitude of a particle executing SHM is $3 \mathrm{cm}$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: ______ $\mathrm{cm}$.

Choose the correct length $\left(L\right)$ versus square of time period $\left(T_2\right)$ ) graph for a simple pendulum executing simple harmonic motion.

A block is fastened to a horizontal spring. The block is pulled to a distance $x=10 \mathrm{cm}$ from its equilibrium position (at $x=0$) on a frictionless surface from rest. The energy of the block at $x=5 \mathrm{cm}$ is $0.25 \mathrm{J}$. The spring constant of the spring is ______ $\mathrm{N} {\mathrm{m}}^{-1}$.