When a body of mass $1.0 \,kg$ is suspended from a certain light spring hanging vertically, its length increases by $5 \,cm$. By suspending a $2.0 \,kg$ block to the spring and if the block is pulled through $10 \,cm$ and released, the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 \,m/s^2$)
- A$0.5$
- B$1$
- C$2$
- D$4$
Explore More
Similar Questions
$A$ spring-block system in horizontal oscillation has a time period $T$. If the spring is cut into four equal parts and the block is re-connected with one of the parts,what will be the new time period of oscillation?
Easy
View SolutionIn the given figure,a body of mass $M$ is held between two massless springs on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has a spring constant $k,$ the frequency of oscillation of the given body is:
DifficultJEE MAIN 2021
View Solution$A$ block of mass $m$ attached to one end of a vertical spring produces an extension $x$. If the block is pulled and released,the periodic time of oscillation is:
$A$ body of mass $1 \,kg$ is attached to the lower end of a vertically suspended spring of force constant $600 \,N \,m^{-1}$. If another body of mass $0.5 \,kg$ moving vertically upward hits the suspended body with a velocity $3 \,m \,s^{-1}$ and gets embedded in it, then the frequency of the oscillation is
All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:
Vedclass 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