A block of mass m attached to a massless spri | vedclass

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

Question

(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.When half the mass breaks off

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(B) At the equilibrium position,the velocity of the block is maximum,given by $V_0 = \omega_0 A = \sqrt{\frac{k}{m}} A$.When half the mass breaks off at the equilibrium position,the velocity of the remaining mass $m/2$ remains the same as $V_0$ because there is no impulsive force acting on the system in the horizontal direction.Let the new amplitude be $A'$. The new angular frequency is $\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \omega_0$.Since the velocity at the equilibrium position is $V_0 = \omega' A'$,we have:$\omega_0 A = \omega' A'$$\omega_0 A = (\sqrt{2} \omega_0) A'$$A' = \frac{A}{\sqrt{2}}$.Thus,$f = \frac{1}{\sqrt{2}}$.

Similar Questions

$A$ mass '$m$' attached to a spring oscillates with a period of $3 \ s$. If the mass is increased by $0.6 \ kg$,the period increases by $3 \ s$. The initial mass '$m$' is equal to (in $kg$)

The potential energy of a particle of mass $m$ situated in a unidimensional potential field varies as $U(x) = U_0(1 - \cos ax)$,where $U_0$ and $a$ are constants. The time period of small oscillations of the particle about the mean position is

If a body of mass $0.98\, kg$ is made to oscillate on a spring of force constant $4.84\, N/m$,the angular frequency of the body is ..... $rad/s$. (in $.22$)

Two bodies $A$ and $B$ of equal mass are suspended from two separate massless springs of spring constants $K_1$ and $K_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of $B$ to that of $A$ is

$A$ mass '$m_1$' is suspended from a spring of negligible mass. The spring is pulled slightly in the downward direction and released; the mass performs $S.H.M.$ with a period '$T_1$'. If the mass is increased by '$m_2$',the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module