A mass on a vertical spring begins its motion | vedclass

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

Question

(D) The motion is Simple Harmonic Motion $(SHM)$ between $y = 0 \ cm$ and $y = 10 \ cm$. The mean position is at $y = 5 \ cm$.The amplitude of the osc

Vedclass · Accepted Answer

$F = \frac{8}{25}(5 - y)$

Vedclass · Answer

(D) The motion is Simple Harmonic Motion $(SHM)$ between $y = 0 \ cm$ and $y = 10 \ cm$. The mean position is at $y = 5 \ cm$.The amplitude of the oscillation is $A = 5 \ cm = 0.05 \ m$.From the graph,the maximum kinetic energy is $KE_{max} = 4 \ J$ at the mean position $(y = 5 \ cm)$.The formula for maximum kinetic energy in $SHM$ is $KE_{max} = \frac{1}{2} k A^2$,where $k = m\omega^2$.Substituting the values: $4 = \frac{1}{2} k (0.05)^2$.$4 = \frac{1}{2} k (0.0025) \implies k = \frac{8}{0.0025} = 3200 \ N/m$.The net force in $SHM$ is given by $F = -k(y - y_{mean})$,where $y_{mean} = 5 \ cm = 0.05 \ m$.$F = -3200(y - 0.05) = -3200y + 160$.However,looking at the options provided in terms of $cm$,we use $k_{eff} = \frac{2 KE_{max}}{A^2} = \frac{2 	imes 4}{5^2} = \frac{8}{25} \ N/cm$.Thus,the net force is $F = -\frac{8}{25}(y - 5) = \frac{8}{25}(5 - y)$.

$A$ mass on a vertical spring begins its motion at rest at $y = 0 \ cm$. It reaches a maximum height of $y = 10 \ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy $(KE)$ versus position is given below. Net force on the mass varies with $y$ as

Similar Questions

$A$ block of mass $m$ hangs from three springs having the same spring constant $k$. If the mass is slightly displaced downwards,the time period of oscillation will be

Two springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

$A$ system consisting of three masses attached to a spring is in equilibrium. When the $700\,g$ mass is removed,the system oscillates with a period of $3\,s$. When the $500\,g$ mass is also removed,what will be the new time period of the system? (in $\text{seconds}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module