Obtain the ratio of maximum acceleration and maximum velocity of a $SHM$ particle.
(D) For a particle executing $SHM$ with amplitude $A$ and angular frequency $\omega$,the displacement is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum velocity is $v_{max} = A\omega$.
The acceleration is $a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$. The maximum acceleration is $a_{max} = A\omega^2$.
The ratio of maximum acceleration to maximum velocity is $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum velocity is $v_{max} = A\omega$.
The acceleration is $a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$. The maximum acceleration is $a_{max} = A\omega^2$.
The ratio of maximum acceleration to maximum velocity is $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.
Explore More
Similar Questions
For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then:
Difficult
View Solution$A$ vertical spring-mass system has the same time period as a simple pendulum undergoing small oscillations. Now, both of them are placed in an elevator moving downwards with an acceleration $a = 5 \,m/s^2$. The ratio of the time period of the spring-mass system to the time period of the pendulum is (Assume, acceleration due to gravity, $g = 10 \,m/s^2$)
Consider two identical cylinders (each of mass $m$,density $\rho_0$,horizontal cross-section area $s$) in equilibrium,partially submerged in two containers filled with liquids of densities $\rho_1$ and $\rho_2$ as shown in the figure. Find the period of small oscillations of this system about its equilibrium. Neglect the changes in the level of liquids in the containers. Neglect the mass of the strings. The acceleration due to gravity is $g$. ($v$ is the volume of each block).
The figure shows a graph between velocity $(v)$ and displacement $(x)$ from the mean position of a particle performing $SHM$.
In the figure shown,there is friction between the blocks $P$ and $Q$,but the contact between the block $Q$ and the lower surface is frictionless. Initially,the block $Q$ with block $P$ over it lies at $x=0$,with the spring at its natural length. The block $Q$ is pulled to the right and then released. As the spring-block system undergoes $S.H.M.$ with amplitude $A$,the block $P$ tends to slip over $Q$. $P$ is more likely to slip at:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo