Question

A particle with mass m moves in accordance to the equation $F = - amr$ where $a$ is a constant, $r$ is the radius vector. Also, $r = {r_0}\hat i$ and $v = {v_0}\hat j$ at $t = 0$ . Describe the trajectory of the particle.
A. ${(\dfrac{x}{{{r_0}}})^2} + a{(\dfrac{y}{{{v_0}}})^2} = 1$
B. ${(\dfrac{x}{{{r_0}}})^2} + a{(\dfrac{y}{{{v_0}}})^2} = 0$
C. ${(\dfrac{x}{{{r_0}}})^2} + a{(\dfrac{y}{{{v_0}}})^2} = \dfrac{1}{a}$
D. None of these

Answer 1

Hint: In this question, first we shall resolve the force and the radial vector into the x and y components. Then by using the equation given in the question, we shall obtain a relation between the acceleration and the distance moved by the body. We can conclude from that relation that the body is undergoing simple harmonic motion. Then by using the standard equation for simple harmonic motion, we will obtain the velocities in x and y directions respectively. From the expressions of velocities, after using the initial conditions provided in the question, we will get the answer.

Formula used:
The equation of simple harmonic motion is,
$x = {x_0}\sin (\omega \,t + \theta )$
where $x$ is the displacement from the mean position, ${x_0}$ is the amplitude, $\omega $ is the angular frequency, $t$ is the time and $\theta $ is the phase angle.

Complete step by step answer:
 It is given that $F = - amr$ where a is a constant, r is the radius vector.
Resolving the force and radius vector into their components we get,
${F_x}\hat i + {F_y}\hat j = - am(x\hat i + y\hat j)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(1)$
We know that $F = ma$ where m is the mass of the body and a is the acceleration.
Also, we know that the acceleration is the double derivative of the distance with respect to time $a = \dfrac{{{d^2}r}}{{d{t^2}}}$
Hence, the force can be written as $F = m\dfrac{{{d^2}r}}{{d{t^2}}}$

Substituting in equation (1) we get,
$m\dfrac{{{d^2}x}}{{d{t^2}}}\hat i + m\dfrac{{{d^2}y}}{{d{t^2}}}\hat j = - am(x\hat i + y\hat j)$
Cancelling out the common terms we get,
$\dfrac{{{d^2}x}}{{d{t^2}}}\hat i + \dfrac{{{d^2}y}}{{d{t^2}}}\hat j = - a(x\hat i + y\hat j)$
Equating like terms on both sides we have,
$\dfrac{{{d^2}x}}{{d{t^2}}} = - ax\,\,\,\,\,\,\,\,\,\,\,\,\,........(2)$ and $\dfrac{{{d^2}y}}{{d{t^2}}} = - ay\,\,\,\,\,\,\,\,\,\,\,\,\,\,........(3)$
Since the acceleration is directly proportional to the distance moved, it is a case of simple harmonic motion.

In $x$ direction, for a simple harmonic motion, $a = - \omega {\,^2}x$
Comparing with equation (2)
$\omega {\,^2} = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,........(4)$
The equation of simple harmonic motion is $x = {x_0}\sin (\omega \,t + \theta )$
Differentiating once with respect to t,
$\dfrac{{dx}}{{dt}} = {x_0}\omega \,\cos (\omega \,t + \theta )\,\,\,\,\,\,\,\,\,\,.......(5)$
Given that $v = {v_0}\hat j$ at $t = 0$ . This means ${v_x} = 0$
Substituting in the above equation, we get
$0 = {x_0}\omega \,\cos (\theta )\,$
$ \Rightarrow \cos \theta = 0$
$ \Rightarrow \theta = {90^0}$

Substituting in the equation of simple harmonic motion,
$x = {x_0}\sin (\omega \,t + {90^0})$
$ \Rightarrow x = {x_0}\cos \omega \,t$
Now given that $r = {r_0}\hat i$ at $t = 0$
$x = {r_0}\cos \omega \,t\,\,\,\,\,\,\,\,\,\,\,.........(6)$
In y direction,
The equation of simple harmonic motion is $y = {y_0}\sin (\omega \,t + \phi )$
At $t = 0$
$y = {y_0}\sin \omega \,t$
Differentiating once with respect to t,
$\dfrac{{dy}}{{dt}} = {y_0}\omega \,\cos \omega \,t\,\,\,\,\,\,\,\,\,.......(7)$

Given that $v = {v_0}\hat j$ at $t = 0$
${v_0} = {y_0}\omega \,\cos 0\,$
$ \Rightarrow {v_0} = {y_0}\omega \,$
Substituting this in equation of simple harmonic motion,
$y = \dfrac{{{v_0}}}{\omega }\sin \omega \,t\,\,\,\,\,\,\,\,\,\,\,\,.........(8)$
From (6) and (8)
After squaring and adding, we get,
${(\dfrac{x}{{{r_0}}})^2} + {(\dfrac{{\omega y}}{{{v_0}}})^2} = {\sin ^2}\omega \,t + {\cos ^2}\omega \,t$
From (4) we know that $\omega {\,^2} = a\,$
So, the equation becomes,
$\therefore {(\dfrac{x}{{{r_0}}})^2} + a{(\dfrac{y}{{{v_0}}})^2} = 1$

Hence, option A is the correct answer.

Note:The restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position. All the simple harmonic motions are oscillatory and also periodic but not all oscillatory motions are SHM. Such a kind of motion is expressed with the help of sinusoids.