In the arrangement given in the figure,if the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When the block of mass $m$ is displaced by a small distance $x$ towards one side,one spring gets compressed by $x$ and the other gets stretched by

Vedclass · Accepted Answer

$n = \frac{1}{2\pi}\sqrt{\frac{k_1 + k_2}{m}}$

Vedclass · Answer

(B) When the block of mass $m$ is displaced by a small distance $x$ towards one side,one spring gets compressed by $x$ and the other gets stretched by $x$.Both springs exert a restoring force in the same direction,opposing the displacement.The restoring force is $F = -(k_1 x + k_2 x) = -(k_1 + k_2)x$.This is equivalent to a single spring with an effective spring constant $k_{eff} = k_1 + k_2$.The frequency of oscillation $n$ is given by $n = \frac{1}{2\pi}\sqrt{\frac{k_{eff}}{m}}$.Substituting $k_{eff}$,we get $n = \frac{1}{2\pi}\sqrt{\frac{k_1 + k_2}{m}}$.

In the arrangement given in the figure,if the block of mass $m$ is displaced,the frequency is given by:

Similar Questions

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is . . . . . . .

$A$ particle of mass $m$ is attached to $3$ springs $A$,$B$ and $C$ of equal force constant $K$ as shown in the figure. If the particle is pushed slightly along the line of spring $C$ and released,find the period of oscillation.

In a spring-block system as shown in the figure,if the spring constant $K = 9 \pi^2 \ Nm^{-1}$,then the time period of oscillation is (in $s$)

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module