A block is in simple harmonic motion (S.H.M) | vedclass

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(D) The equation for position in $S.H.M$ is given by $x(t) = A \cos(\omega t + \phi)$, where $A = 5$ is the amplitude.The total mechanical energy $E$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) The equation for position in $S.H.M$ is given by $x(t) = A \cos(\omega t + \phi)$, where $A = 5$ is the amplitude.The total mechanical energy $E$ of an $S.H.M$ system is given by $E = \frac{1}{2} k A^2 = 100 \ J$.The potential energy $U$ at any time $t$ is given by $U = \frac{1}{2} k x^2$.At $t = 0$, the position is $x(0) = 5 \cos\left(0 + \frac{\pi}{4}ight) = 5 \cos\left(\frac{\pi}{4}ight) = 5 	imes \frac{1}{\sqrt{2}}$.Substituting $x(0)$ into the potential energy formula:$U(0) = \frac{1}{2} k \left(5 	imes \frac{1}{\sqrt{2}}ight)^2 = \frac{1}{2} k \left(\frac{25}{2}ight) = \frac{1}{4} k (25)$.Since $E = \frac{1}{2} k A^2 = \frac{1}{2} k (5)^2 = \frac{25}{2} k = 100 \ J$, we find $k = \frac{200}{25} = 8 \ J/m^2$.Substituting $k = 8$ into the expression for $U(0)$:$U(0) = \frac{1}{4} 	imes 8 	imes 25 = 2 	imes 25 = 50 \ J$.

$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$, then the potential energy at time $t = 0$ is: (in $J$)

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