$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:
- A$T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$
- B$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$
- C$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$
- D$T = 2\pi \sqrt{\frac{m K_1 K_2}{K_1 + K_2}}$
Explore More
Similar Questions
$A$ uniform circular disc of mass $12 \ kg$ is held by two identical springs. When the disc is slightly pressed down and released,it executes $S.H.M.$ of period $2 \ s$. The force constant of each spring is (nearly) (Take $\pi^2=10$) (in $Nm^{-1}$)
$A$ solid sphere of mass $M$ and radius $R$ is attached to a spring of negligible mass kept on a horizontal plane such that it can roll without slipping. The sphere is made to execute $SHM$ by stretching it through a distance and releasing it. The time period of such oscillation is ($K=$ spring constant).
The potential energy of a particle of mass $0.1\,kg,$ moving along the $x-$axis,is given by $U = 5x(x-4)\,J,$ where $x$ is in metres. It can be concluded that
$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)
MediumNEET 2021
View Solution$A$ mass $m$ is suspended separately by two different springs,and the time periods are $t_1$ and $t_2$ respectively. If it is connected by both springs in parallel as shown in the figure,then the time period is $t_0$. The correct relation is:
Difficult
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