Q.Three girls skating on a circular ice ground of radius $200\ \text{m}$ start from a point $P$ on the edge of the ground and reach a point $Q$ that is diametrically opposite to $P$, each following a different path across the ground. The first girl follows a curved path that bulges out to the left, the second girl travels along the straight diameter directly from $P$ to $Q$, and the third girl follows a wavy (zig-zag) path that bulges out to the right, as shown below.

Concept & Intuition

When two projectiles are launched from the same point with the same speed but at complementary angles ($\theta$ and $90^\circ - \theta$), their trajectories are symmetric. At the instant each reaches its maximum height, the vertical velocity is zero, and the horizontal velocity remains constant ($u\cos\theta$ for the first, $u\sin\theta$ for the second). The horizontal distance each has traveled from the launch point at that moment is exactly half its total range. The total separation between them is the sum of these two half‑ranges, because they are thrown in opposite directions.

1. Time to reach maximum height For a projectile launched at angle $\theta$ with speed $u$, the vertical component is $u\sin\theta$. At the top, $v_y = 0$, so

2. Horizontal distance covered in that time Horizontal velocity is constant: $u\cos\theta$. So the horizontal distance from the launch point at maximum height is

3. For the second projectile (angle $90^\circ - \theta$) Its horizontal velocity is $u\cos(90^\circ - \theta) = u\sin\theta$, and its time to maximum height is

Hence its horizontal distance from the launch point at maximum height is

4. Separation when both are at their maximum heights Since they are thrown in opposite directions, the total horizontal distance between them is

Using the identity $2\sin\theta\cos\theta = \sin 2\theta$, we get

The range of a projectile depends on both the launch speed and the angle. Maximum range is achieved at $45°$ (or $\frac{\pi}{4}$ radians), and knowing this maximum lets us work backward to find the launch speed. Once we have the speed, we can calculate the range at any other angle.

The range formula is

where $u$ is the launch speed, $\theta$ is the angle of projection, and $g$ is acceleration due to gravity.

At maximum range, $\theta = \frac{\pi}{4}$, so $\sin(2\theta) = \sin\left(\frac{\pi}{2}\right) = 1$. This gives

This tells us that $\frac{u^2}{g} = 80$ m, which is the key quantity we need.

Now we find the range when the projectile is launched at $\theta = \frac{\pi}{12}$ (which is $15°$):

1. Calculate the angle term: We need $\sin(2\theta)$ where $\theta = \frac{\pi}{12}$.

$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ …

The motion of a particle executing Simple Harmonic Motion (SHM) can be described by its displacement from the equilibrium position as a function of time. The key to solving this problem lies in correctly interpreting the conditions given for the displacement and velocity of the two particles at the moment they pass each other.

Concept and Intuition

A particle undergoing SHM has a displacement $x(t)$ given by:

where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is time, and $\phi$ is the initial phase constant. The term $(\omega t + \phi)$ is the instantaneous phase of the particle.

The velocity $v(t)$ of the particle is the time derivative of its displacement:

The problem states that both particles have the same frequency ($\omega$) and same amplitude ($A$). Let their initial phase constants be $\phi_1$ and $\phi_2$. We are looking for the phase difference, which is $|\phi_1 - \phi_2|$.

When the particles "pass each other," it means their displacements are equal at that instant. The condition "moving in opposite direction" means their velocities must have opposite signs at that same instant. The specific displacement is given as "half their amplitude," meaning $x = A/2$.

We will set up the equations for displacement and velocity for both particles, apply these conditions, and then solve for the phase difference.

Step-by-Step Derivation

1. Set up the equations for displacement and velocity: Let the displacement equations for the two particles be:

Their corresponding velocity equations are:

We are interested in the phase difference, $\Delta\phi = |\phi_1 - \phi_2|$.

2. Apply the displacement condition:

At the moment they pass each other, their displacements are equal and half their amplitude. Let this time be $t_0$.

Substituting this into the displacement equations:

Let $\theta_1 = \omega t_0 + \phi_1$ and $\theta_2 = \omega t_0 + \phi_2$. Then we have:

This means that $\theta_1$ and $\theta_2$ can be $\pi/6$ (first quadrant) or $5\pi/6$ (second quadrant), or angles coterminal with these.

3. Apply the velocity condition:

At the same instant $t_0$, the particles are moving in opposite directions. This means their velocities have opposite signs:

Substituting the velocity equations:

Dividing by $A\omega$ (since $A \neq 0, \omega \neq 0$):

4. Combine conditions to find the phase difference:

   We have two conditions for $\theta_1$ and $\theta_2$:

   1. $\sin(\theta_1) = \frac{1}{2}$ and $\sin(\theta_2) = \frac{1}{2}$
   2. $\cos(\theta_1) = -\cos(\theta_2)$ From condition (i), $\theta_1$ and $\theta_2$ must be angles whose sine is $1/2$. These are typically $\pi/6$ or $5\pi/6$ (within the range $[0, 2\pi)$). Let's consider the possible values for $\theta_1$:
   • Case 1: If $\theta_1 = \pi/6$ Then $\cos(\theta_1) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$. From condition (ii), $\cos(\theta_2) = -\cos(\theta_1) = -\frac{\sqrt{3}}{2}$. …