$A$ 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

  • A
    $\frac{5}{\pi} \,Hz$
  • B
    $\frac{10}{\pi} \,Hz$
  • C
    $\frac{\pi}{5} \,Hz$
  • D
    $\pi \,Hz$

Explore More

Similar Questions

In the figure,$S_1$ and $S_2$ are identical springs. The oscillation frequency of the mass $m$ is $f$. If one spring is removed,the frequency will become

Two blocks of masses $m$ and $M$ $(M > m)$ are placed on a frictionless table as shown in the figure. $A$ massless spring with spring constant $k$ is attached to the lower block. If the system is slightly displaced and released,then ($\mu =$ coefficient of friction between the two blocks):
$(A)$ The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{M + m}{k}}$
$(B)$ The acceleration of the blocks is $a = \frac{kx}{M + m}$ ($x =$ displacement of the blocks from the mean position)
$(C)$ The magnitude of the frictional force on the upper block is $f = \frac{mkx}{M + m}$
$(D)$ The maximum amplitude of the upper block,if it does not slip,is $A = \frac{\mu mg(M + m)}{mk} = \frac{\mu g(M + m)}{k}$ (Wait,let's re-evaluate: $f_{max} = \mu mg$. Since $f = ma = m \cdot \frac{kx}{M+m}$,at max amplitude $A$,$m \cdot \frac{kA}{M+m} = \mu mg \implies A = \frac{\mu g(M+m)}{k}$)
$(E)$ Maximum frictional force can be $\mu mg$.
Choose the correct answer from the options given below.

Two identical springs of constant $K$ are connected in series and parallel as shown in the figure. $A$ mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $k$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system; the amplitude of vibration is

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal,the ratio of the amplitude of $M$ to that of $N$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo