The variation of potential energy of a harmon | vedclass

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

Question

(B) From the given graph,the potential energy $U$ varies with displacement $y$. The potential energy of a harmonic oscillator is given by $U(y) = U_0

Vedclass · Accepted Answer

$150 	ext{ N/m}$

Vedclass · Answer

(B) From the given graph,the potential energy $U$ varies with displacement $y$. The potential energy of a harmonic oscillator is given by $U(y) = U_0 + \frac{1}{2}ky^2$,where $U_0$ is the minimum potential energy at the equilibrium position $(y=0)$.From the graph,the minimum potential energy at $y=0$ is $U_0 = 0.01 	ext{ J}$.The maximum potential energy at the extreme position $y = 20 	ext{ mm} = 0.02 	ext{ m}$ is $U_{max} = 0.04 	ext{ J}$.Therefore,the change in potential energy from equilibrium to the extreme position is $\Delta U = U_{max} - U_0 = 0.04 	ext{ J} - 0.01 	ext{ J} = 0.03 	ext{ J}$.This change in potential energy is equal to the elastic potential energy stored in the spring,which is $\frac{1}{2}kA^2$,where $A$ is the amplitude.$0.03 = \frac{1}{2} 	imes k 	imes (0.02)^2$$0.03 = \frac{1}{2} 	imes k 	imes 0.0004$$0.03 = k 	imes 0.0002$$k = \frac{0.03}{0.0002} = \frac{300}{2} = 150 	ext{ N/m}$.

The variation of potential energy of a harmonic oscillator is as shown in the figure. The spring constant is

Similar Questions

In the given arrangement,the spring constant $k$ has a value of $2\,N\,m^{-1}$,mass $M = 3\,kg$,and mass $m = 1\,kg$. Mass $M$ is in contact with a smooth surface. The coefficient of friction between the two blocks is $0.1$. The time period of $SHM$ executed by the system is

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$,respectively. If the maximum velocities during oscillations are equal,the ratio of amplitudes of $A$ and $B$ is

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by $m$,the time period becomes $5T/3$. Then the ratio of $m/M$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module