Question

A particle of mass 10 gm moves in a field where potential energy per unit mass is given by expression $v = 8 \times {10^4}{x^2}erg/gm$. If the total energy of the particle is $8 \times {10^7}erg$then the relation between x and t is:
$[\phi = {\rm{constant]}}$
$A.\,\,x = 10\sin (400t + \phi )cm\\\ B.\,\,x = \sin (400t + \phi )cm\\\ C.\,\,x = 10\sin (4t + \phi )cm\\\ D.\,\,x = 100\sin (400t + \phi )cm$

Answer 1

In this type of question where a quantity varies with a variable, we have to find a suitable relation and perform the necessary calculus operations. In the question we have been given the expression of potential energy per unit mass which changes with position and we have to properties like angular velocity and amplitude between the relation of x and t to get the answer.

Complete step by step answer:
We are given the equation of potential energy per unit mass and we have to calculate the potential energy function:We know that,
${\rm{potential energy = potential}} \times {\rm{mass}}$
Therefore, the function of potential energy:
${\rm{Potential energy }} =8 \times {10^4}{x^2} \times 10\\\ {\rm{Potential energy }} =8 \times {10^5}{x^2}erg$

When a particle moves in such a field then the motion it produces is S.H.M.
Hence to find the angular velocity we have to first find the force.
We know that force is given by:
$F = \dfrac{{ - dU}}{{dx}}$, where U is the potential energy.
Putting the function of U and differentiating, we get:
$F = - \dfrac{{dU}}{{dx}}\\\ \Rightarrow F = - \dfrac{{d(8 \times {{10}^5} \times {x^2})}}{{dx}}\\\ \Rightarrow F = - 8 \times {10^5}\dfrac{{d({x^2})}}{{dx}}\\\ \Rightarrow F = - 8 \times {10^5}(2x)\\\ \Rightarrow F = - (16 \times {10^5})x$

Now we have the function of force and we can use that to calculate the angular velocity $(\omega )$.
We know that $F = ma$, therefore we can find the acceleration as:
$a = \dfrac{F}{m}\\\ \Rightarrow a = \dfrac{{ - 16 \times {{10}^5} \times x}}{{10}}\\\ \Rightarrow a = - 16 \times {10^4} \times x$
The negative sign is only for direction and we will not use it for calculations.
Now we have the equation of acceleration and using that we can find angular velocity $(\omega )$.
Acceleration in SHM is given as:
$a = {\omega ^2}x$
Now we will put the equation of acceleration in the above relation:
$a = {\omega ^2}x\\\ {\omega ^2}x = a\\\ \Rightarrow {\omega ^2}x = 16 \times {10^4} \times x\\\ \Rightarrow {\omega ^2} = 16 \times {10^4}\\\ \Rightarrow \omega = \sqrt {16 \times {{10}^4}} \\\ \Rightarrow \omega = 400$

Now we know that option cannot be the answer.Now we will calculate the Amplitude.We will do this by taking the relation of Energy in SHM:
${\text{Total energy in SHM }} = \dfrac{1}{2} \times m \times {\omega ^2}{A^2}\\\ \Rightarrow 8 \times {10^7} = \dfrac{1}{2} \times 10 \times 16 \times {10^4} \times {A^2}\\\ \Rightarrow 8\times {10^7} = 10 \times 8 \times {10^4} \times {A^2}\\\ \Rightarrow {A^2} = 100\\\ \therefore A = 10cm$
Therefore, the relation can now be easily distinguished as we have to check the option where the value of amplitude and the angular velocity is matching to the ones we calculated.

Hence, the correct option is A.

Note: In such types of questions, knowledge of calculus is very important. Also, students should know the relations between quantities as we have to derive a suitable relation so as to apply calculus operations. The conversion of units is very important as well.