When a mass $m$ is hung from the lower end of a spring of negligible mass,an extension $x$ is produced in the spring. The time period of oscillation is
- A$T = 2\pi \sqrt {\frac{x}{g}}$
- B$T = 2\pi \sqrt {\frac{gx}{m}}$
- C$T = 2\pi \sqrt {\frac{m}{k}}$
- D$T = 2\pi \sqrt {\frac{2x}{g}}$
Explore More
Similar Questions
The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is
DifficultIIT 2009
View SolutionThe equivalent spring constant of the system as shown in the figure will be:
Difficult
View SolutionWhen the mass attached to a spring is increased from $4 \ kg$ to $9 \ kg$,the time period of oscillation increases by $0.2 \pi \ s$. Then the spring constant of the spring is (in $N \ m^{-1}$)
Two identical particles each of mass $m$ are interconnected by a light spring of stiffness $k$. The time period for small oscillations is equal to:
Difficult
View Solution$A$ block of mass $M = 5 \ kg$ is suspended from one end of a spring. The spring extends by $l = 0.1 \ m$ due to the mass of the block. The block is given an upward velocity of $v = 2 \ m/s$. What is the maximum height $h$ (in $m$) reached by the block? $(g = 10 \ m/s^2)$
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