A particle of mass 1 , text{kg} is hanging fr | vedclass

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

Question

(C) For a particle in simple harmonic motion,the total energy $E$ is the sum of kinetic energy $(KE)$ and potential energy $(PE)$.$E = KE + PE$Given t

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) For a particle in simple harmonic motion,the total energy $E$ is the sum of kinetic energy $(KE)$ and potential energy $(PE)$.$E = KE + PE$Given that $KE = PE$,we have $E = 2 \cdot PE$ or $E = 2 \cdot KE$.The potential energy at displacement $y$ is $PE = \frac{1}{2} k y^2$,and the total energy is $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.Setting $KE = PE$ implies $PE = \frac{1}{2} E$,so $\frac{1}{2} k y^2 = \frac{1}{2} (\frac{1}{2} k A^2)$.This simplifies to $y^2 = \frac{A^2}{2}$,or $y = \frac{A}{\sqrt{2}}$.Using the equation of motion $y = A \sin(\omega t)$,we get $\frac{A}{\sqrt{2}} = A \sin(\omega t)$,which means $\sin(\omega t) = \frac{1}{\sqrt{2}}$.Thus,$\omega t = \frac{\pi}{4}$.Since $\omega = \frac{2\pi}{T}$,we have $(\frac{2\pi}{T}) t = \frac{\pi}{4}$.Solving for $t$,we get $t = \frac{T}{8}$.Comparing this with $\frac{T}{x}$,we find $x = 8$.

Similar Questions

$A$ body of mass $4 \ kg$ attached to a spring of force constant $64 \ N \ m^{-1}$ executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is

Two springs of force constants $300 \, N/m$ (Spring $A$) and $400 \, N/m$ (Spring $B$) are joined together in series. The combination is compressed by $8.75 \, cm$. The ratio of energy stored in $A$ and $B$ is $E_A / E_B$. Then $E_A / E_B$ is equal to:

Two springs,of force constants $k_1$ and $k_2$,are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

Two particles $A$ and $B$ of masses $m$ and $2m$ are suspended from massless springs of force constants $K_1$ and $K_2$. During their oscillation,if their maximum velocities are equal,then the ratio of amplitudes of $A$ and $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module