Question

A particle of mass 10 gm is moving with P.E $v =0.08 x^2$ Joule, where x is in m. Obtain the time displacement for the particle if its total energy is $8\times {10^{-4}}$ Joule.

Answer 1

Hint: First we are calculating the force and then the spring constant. The frequency of simple harmonic motion like a mass on a spring is determined by the mass m and the stiffness of the spring expressed in terms of a spring constant k.

Complete step by step answer:
Given that the potential energy is,$0.08 x^2$ Joule and the total energy is given as $8\times {10^{-4}}$ Joule. We can find the force $F$ by differentiating the potential energy with respect to x. So we have,
\[F = 0.16x\;{\rm{N}}\].
As we know, the spring constant is, k = Fx. Therefore, we have,
$k = 0.16$
Also we have the energy in simple harmonic motion as,
$E = \dfrac{1}{2}k{A^2}$
Here, A is the amplitude. Therefore on substituting the value of k and the total energy in the above equation, we have,
$
8 \times {10^{ - 4}}{\rm{Joule}} = \dfrac{1}{2} \times 0.16 \times {A^2}\\
\Rightarrow\;\;\;\;\;\;\;\;{A^2} = \dfrac{{2 \times 8 \times {{10}^{ - 4}}{\rm{Joule}}}}{{0.16}}\\
\Rightarrow A = 10\;{\rm{m}}
$
Thus, we have an amplitude of 10 m. Therefore now we can calculate the angular frequency by using the equation,
$\omega = \sqrt {\dfrac{k}{m}} $
We have the value of k as 0.16 and the mass m is given as 10 gm. On converting it into kilograms we have mass m as 10 x 10-3 kg. Thus on substituting we have,
$\omega = \sqrt {\dfrac{{0.16}}{{10 \times {{10}^{ - 3}}}}} = 4$
Thus the position or displacement can now be expressed in terms of time as,
$\therefore x = 10(\sin 4t + \phi )m$

Therefore the time displacement for the particle of mass 10 gm moving with P.E $v =0.08
x^2$ Joule and total energy is $8\times {10^{-4}}$ Joule is $x = 10(\sin 4t + \phi )m$.

Note: Here we are taking the phase as constant The relationship between amplitude and the spring constant is then employed to find the angular frequency.Spring is a tool used daily by many of us and their inertia is frequently neglected by assuming it as massless. It’s an extremely casual activity that a Spring when strained, undergoes displacement when it is compacted it gets compressed and when it is free it comes to its equilibrium position. This fact tells us that spring exerts an equal as well as an opposite force on a body which compresses or stretches it.