Question

Time Period = 4 Sec., find Time at which they are at Same position for First time., Velocity & disp from MP at that time.

Answer 1

Time = 11/6 s, Displacement from MP = -A * (sqrt(3)-1) / (2sqrt(2)), Velocity Magnitude = Api*(sqrt(3)+1) / (4*sqrt(2))

Answer 2

The problem describes two pendulums undergoing Simple Harmonic Motion (SHM) with the same time period $T = 4$ seconds. We need to find the time when they are at the same position for the first time, their common displacement from the mean position (MP) at that time, and their velocities.

First, let's find the angular frequency $\omega$: $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}$ rad/s.

Let the amplitude of oscillation for both pendulums be $A$.

1. Equations of Motion for each pendulum:

• Pendulum 1:
  At $t=0$, it is at the Mean Position ($x_1=0$) and moving to the left (negative velocity).
  The general equation for SHM is $x(t) = A \sin(\omega t + \phi)$.
  At $t=0$, $x_1(0) = A \sin(\phi_1) = 0 \implies \phi_1 = 0$ or $\phi_1 = \pi$.
  The velocity is $v_1(t) = \frac{dx_1}{dt} = A\omega \cos(\omega t + \phi_1)$.
  At $t=0$, $v_1(0) = A\omega \cos(\phi_1)$. Since it's moving left, $v_1(0) < 0$.
  This requires $\cos(\phi_1) < 0$. Thus, $\phi_1 = \pi$.
  So, the equation for Pendulum 1 is:
  $x_1(t) = A \sin(\omega t + \pi) = -A \sin(\omega t)$.

• Pendulum 2:
  At $t=0$, it is at $x_2 = A/2$ and moving to the right (positive velocity).
  Using $x_2(t) = A \sin(\omega t + \phi_2)$:
  At $t=0$, $x_2(0) = A \sin(\phi_2) = A/2 \implies \sin(\phi_2) = 1/2$.
  This implies $\phi_2 = \pi/6$ or $\phi_2 = 5\pi/6$.
  The velocity is $v_2(t) = A\omega \cos(\omega t + \phi_2)$.
  At $t=0$, $v_2(0) = A\omega \cos(\phi_2)$. Since it's moving right, $v_2(0) > 0$.
  This requires $\cos(\phi_2) > 0$. Thus, $\phi_2 = \pi/6$.
  So, the equation for Pendulum 2 is:
  $x_2(t) = A \sin(\omega t + \pi/6)$.

2. Time when they are at the same position for the first time:

Set $x_1(t) = x_2(t)$: $-A \sin(\omega t) = A \sin(\omega t + \pi/6)$ $\sin(\omega t + \pi/6) + \sin(\omega t) = 0$ Using the sum-to-product trigonometric identity $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$: $2 \sin\left(\frac{\omega t + \pi/6 + \omega t}{2}\right) \cos\left(\frac{\omega t + \pi/6 - \omega t}{2}\right) = 0$ $2 \sin\left(\omega t + \frac{\pi}{12}\right) \cos\left(\frac{\pi}{12}\right) = 0$ Since $\cos(\pi/12) \neq 0$, we must have: $\sin\left(\omega t + \frac{\pi}{12}\right) = 0$ For the first positive time $t$, we need the smallest positive value for the argument. $\omega t + \frac{\pi}{12} = \pi$ (since $\omega t + \pi/12 = 0$ would give $t<0$) $\omega t = \pi - \frac{\pi}{12} = \frac{11\pi}{12}$ Substitute $\omega = \pi/2$: $\frac{\pi}{2} t = \frac{11\pi}{12}$ $t = \frac{11\pi}{12} \times \frac{2}{\pi} = \frac{11}{6}$ seconds.

3. Displacement from Mean Position at that time:

Substitute $t = \frac{11}{6}$ s into $x_1(t)$: $x = -A \sin\left(\frac{\pi}{2} \times \frac{11}{6}\right) = -A \sin\left(\frac{11\pi}{12}\right)$ We know that $\sin(\frac{11\pi}{12}) = \sin(\pi - \frac{\pi}{12}) = \sin(\pi/12)$. $\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ $= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$. So, the displacement is: $x = -A \frac{\sqrt{3}-1}{2\sqrt{2}}$.

4. Velocity at that time:

For Pendulum 1: $v_1(t) = -A\omega \cos(\omega t)$ $v_1\left(\frac{11}{6}\right) = -A\left(\frac{\pi}{2}\right) \cos\left(\frac{11\pi}{12}\right)$ We know that $\cos(\frac{11\pi}{12}) = \cos(\pi - \frac{\pi}{12}) = -\cos(\pi/12)$. $\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$ $= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$. So, $v_1\left(\frac{11}{6}\right) = -A\left(\frac{\pi}{2}\right) \left(-\frac{\sqrt{3}+1}{2\sqrt{2}}\right) = A\frac{\pi(\sqrt{3}+1)}{4\sqrt{2}}$.

For Pendulum 2: $v_2(t) = A\omega \cos(\omega t + \pi/6)$ $v_2\left(\frac{11}{6}\right) = A\left(\frac{\pi}{2}\right) \cos\left(\frac{11\pi}{12} + \frac{\pi}{6}\right) = A\left(\frac{\pi}{2}\right) \cos\left(\frac{11\pi + 2\pi}{12}\right) = A\left(\frac{\pi}{2}\right) \cos\left(\frac{13\pi}{12}\right)$ We know that $\cos(\frac{13\pi}{12}) = \cos(\pi + \frac{\pi}{12}) = -\cos(\pi/12) = -\frac{\sqrt{3}+1}{2\sqrt{2}}$. So, $v_2\left(\frac{11}{6}\right) = A\left(\frac{\pi}{2}\right) \left(-\frac{\sqrt{3}+1}{2\sqrt{2}}\right) = -A\frac{\pi(\sqrt{3}+1)}{4\sqrt{2}}$.

At the time they are at the same position, their velocities are equal in magnitude but opposite in direction. The question asks for "Velocity", which typically refers to the magnitude unless specified. So, the magnitude of velocity is $A\frac{\pi(\sqrt{3}+1)}{4\sqrt{2}}$.