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

$A$ body executes $SHM$ under the action of force $F_1$ with time period $T_1$. If the force is changed to $F_2$,it executes $SHM$ with time period $T_2$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period is:

In the situation as shown in the figure,the time period of vertical oscillation of the block for small displacements will be:

$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)

$A$ body at the end of a spring executes $S.H.M.$ with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

$A$ mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_1$ and $K_2$. For the frictionless surface,the time period of oscillation of mass $m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module