Question

The differential equation for linear SHM of a particle of mass $2\;g$ is $\dfrac{{{d^2}x}}{{d{t^2}}} + 16x = 0$. Find the force constant.

Answer 1

To solve this question we need to use the conditions of linear SHM to arrive at the formula for the force acting on the particle. Then equate this formula to Newton’s law of motion. We know the differential equation of the acceleration of the particle. Substitute this and compare it with the differential equation given in the question to arrive at the answer.
Formula used:
The force acting on a particle moving in SHM is given as,
$F = - kx$
From Newton’s law of motion, we know that,
$F = ma$

Complete step by step solution:
We know that the force acting on a particle moving in SHM is given as,
$F = - kx$ ……….$(1)$
where, $k$ is the force constant that we have to find out from the question, and
$x$ is the displacement of the particle.
From Newton’s law of motion, we know that,
$F = ma$
Again we know that acceleration is the rate of change of velocity.
$\therefore a = \dfrac{{dv}}{{dt}}$
Again we know that velocity is the rate of change of displacement,
$\therefore v = \dfrac{{dx}}{{dt}}$
Therefore we can say that,
$a = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right)$
$\Rightarrow a = \dfrac{{{d^2}x}}{{d{t^2}}}$
Or, $F = m\dfrac{{{d^2}x}}{{d{t^2}}}$ ……….$(2)$
where, $m$ is the mass of the particle, and
$a$ is the acceleration of the particle.
Upon equating equations $(1)$ and $(2)$ we get,
$F = - kx = m\dfrac{{{d^2}x}}{{d{t^2}}}$
$\Rightarrow m\dfrac{{{d^2}x}}{{d{t^2}}} + kx = 0$
Dividing the above equation by $m$ we get,
$\dfrac{{{d^2}x}}{{d{t^2}}} + \dfrac{k}{m}x = 0$
$\Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + \omega x = 0$ ……….$(3)$
where $\omega = \dfrac{k}{m}$
Equation $(3)$ is called the general differential equation of a particle executing SHM.
In the question, the differential equation for linear SHM of a particle is given as,
$\dfrac{{{d^2}x}}{{d{t^2}}} + 16x = 0$
Comparing this equation with equation $(3)$ we can see that,
$\omega = 16$
Again we have $\omega = \dfrac{k}{m}$,
or $k = \omega m$
In the question, it is given that $m = 2g$.
Therefore we have, $k = 16 \times 2 = 32N/m$

The force constant of the differential equation for linear SHM of the given particle is $32N/m$.

Note: The property of a particle executing linear SHM is that the force acting on the particle is proportional to the displacement, or the direction from the fixed point. The motion is always directed towards some fixed point in its path which is the opposite of the direction of motion of the particle. Hence the force $F = - kx$ is negative.