Question

Two bodies are executing S.H.M. such that their amplitude is A and time period T when one of the blocks is at right extreme another block is at +$\frac{A}{2}$ moving towards left. Find the time after which both the blocks will have same position from equilibrium position.

• A. T/12
• B. T/6
• C. 5T/12
• D. T/3

Answer 1

Answer: (C)

5T/12

Answer 2

To solve this problem, we need to set up the equations of motion for both blocks undergoing Simple Harmonic Motion (S.H.M.) and then find the time when their positions are equal.

Let the angular frequency be $\omega = \frac{2\pi}{T}$. The general equation for S.H.M. can be written as $x(t) = A \cos(\omega t + \phi)$, where $A$ is the amplitude, $T$ is the time period, and $\phi$ is the initial phase.

1. Equation of motion for Body 1: At $t=0$, Body 1 is at the right extreme position, $x_1(0) = +A$. Using $x_1(t) = A \cos(\omega t + \phi_1)$: $A = A \cos(\phi_1) \Rightarrow \cos(\phi_1) = 1$. The simplest choice for $\phi_1$ is $0$. So, the equation of motion for Body 1 is: $x_1(t) = A \cos(\omega t)$

2. Equation of motion for Body 2: At $t=0$, Body 2 is at $x_2(0) = +\frac{A}{2}$ and moving towards the left. Using $x_2(t) = A \cos(\omega t + \phi_2)$: $\frac{A}{2} = A \cos(\phi_2) \Rightarrow \cos(\phi_2) = \frac{1}{2}$. This gives two possible values for $\phi_2$ in the range $[0, 2\pi)$: $\phi_2 = \frac{\pi}{3}$ or $\phi_2 = \frac{5\pi}{3}$.

Now, we use the velocity condition. The velocity of an SHM is $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$. At $t=0$, $v_2(0) = -A\omega \sin(\phi_2)$. Since the block is moving towards the left, its initial velocity $v_2(0)$ must be negative. So, $-A\omega \sin(\phi_2) < 0 \Rightarrow \sin(\phi_2) > 0$.

Let's check the possible values of $\phi_2$:

• If $\phi_2 = \frac{\pi}{3}$, then $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} > 0$. This is consistent with the condition.
• If $\phi_2 = \frac{5\pi}{3}$, then $\sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2} < 0$. This is not consistent.

Therefore, the correct initial phase for Body 2 is $\phi_2 = \frac{\pi}{3}$. The equation of motion for Body 2 is: $x_2(t) = A \cos\left(\omega t + \frac{\pi}{3}\right)$

3. Finding the time when both blocks have the same position: We need to find the smallest positive time $t$ such that $x_1(t) = x_2(t)$. $A \cos(\omega t) = A \cos\left(\omega t + \frac{\pi}{3}\right)$ Since $A \neq 0$, we can divide by $A$: $\cos(\omega t) = \cos\left(\omega t + \frac{\pi}{3}\right)$ For $\cos(X) = \cos(Y)$, the general solution is $X = \pm Y + 2n\pi$, where $n$ is an integer. Let $X = \omega t$ and $Y = \omega t + \frac{\pi}{3}$.

• Case 1: $\omega t = (\omega t + \frac{\pi}{3}) + 2n\pi$ $0 = \frac{\pi}{3} + 2n\pi$ This equation has no integer solution for $n$, so this case does not yield a valid time.

• Case 2: $\omega t = -(\omega t + \frac{\pi}{3}) + 2n\pi$ $\omega t = -\omega t - \frac{\pi}{3} + 2n\pi$ $2\omega t = -\frac{\pi}{3} + 2n\pi$ $\omega t = -\frac{\pi}{6} + n\pi$

We are looking for the first time after $t=0$, which means we need the smallest positive value for $\omega t$.

• If $n=0$, $\omega t = -\frac{\pi}{6}$ (negative time, not valid).
• If $n=1$, $\omega t = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$. This is the smallest positive value.

Now, substitute $\omega = \frac{2\pi}{T}$: $\frac{2\pi}{T} t = \frac{5\pi}{6}$ $t = \frac{5\pi}{6} \cdot \frac{T}{2\pi}$ $t = \frac{5T}{12}$

This is the time after which both blocks will have the same position from the equilibrium.