Question

Two bodies are executing S.H.M. with same amplitude A and time period T. When one of blocks is located at right extreme and other block is located at equilibrium position moving towards left. Find the time when both the blocks will have same displacement from the equilibrium position.

Answer 1

3T/8

Answer 2

The problem involves two bodies executing Simple Harmonic Motion (S.H.M.) with the same amplitude $A$ and time period $T$. We need to find the time when their displacements from the equilibrium position are equal.

Let the angular frequency be $\omega = \frac{2\pi}{T}$. The general equation for S.H.M. is $x(t) = A \sin(\omega t + \phi)$ or $x(t) = A \cos(\omega t + \phi)$. We choose the form that simplifies the initial conditions.

Body 1: At $t=0$, the first block is at the right extreme position, i.e., $x_1(0) = +A$. 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: $x_1(t) = A \cos(\omega t) = A \cos\left(\frac{2\pi}{T} t\right)$

Body 2: At $t=0$, the second block is at the equilibrium position moving towards the left. This means $x_2(0) = 0$ and its initial velocity $v_2(0) < 0$. Using $x_2(t) = A \sin(\omega t + \phi_2)$: $0 = A \sin(\phi_2) \Rightarrow \sin(\phi_2) = 0$. This implies $\phi_2 = 0$ or $\phi_2 = \pi$. Now, let's check the velocity: $v_2(t) = \frac{dx_2}{dt} = A\omega \cos(\omega t + \phi_2)$. At $t=0$, $v_2(0) = A\omega \cos(\phi_2)$. Since the block is moving towards the left, $v_2(0)$ must be negative. If $\phi_2 = 0$, $v_2(0) = A\omega \cos(0) = A\omega > 0$ (incorrect). If $\phi_2 = \pi$, $v_2(0) = A\omega \cos(\pi) = -A\omega < 0$ (correct). So, the equation of motion for Body 2 is: $x_2(t) = A \sin(\omega t + \pi) = -A \sin(\omega t) = -A \sin\left(\frac{2\pi}{T} t\right)$

Finding the time when both blocks have the same displacement: We need to find $t$ such that $x_1(t) = x_2(t)$. $A \cos\left(\frac{2\pi}{T} t\right) = -A \sin\left(\frac{2\pi}{T} t\right)$ Since $A \neq 0$, we can divide by $A$: $\cos\left(\frac{2\pi}{T} t\right) = -\sin\left(\frac{2\pi}{T} t\right)$ Let $\theta = \frac{2\pi}{T} t$. The equation becomes: $\cos(\theta) = -\sin(\theta)$ Dividing by $\cos(\theta)$ (assuming $\cos(\theta) \neq 0$): $1 = -\tan(\theta) \Rightarrow \tan(\theta) = -1$

We are looking for the first time when this condition is met, so we need the smallest positive value of $t$. For $\tan(\theta) = -1$, the general solution is $\theta = n\pi - \frac{\pi}{4}$, where $n$ is an integer. To find the smallest positive $\theta$: If $n=0$, $\theta = -\frac{\pi}{4}$ (negative, so $t$ would be negative). If $n=1$, $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. This is the smallest positive value for $\theta$.

Now substitute back $\theta = \frac{2\pi}{T} t$: $\frac{2\pi}{T} t = \frac{3\pi}{4}$ $t = \frac{3\pi}{4} \cdot \frac{T}{2\pi}$ $t = \frac{3T}{8}$

At this time $t = \frac{3T}{8}$, the displacement of both blocks will be: $x_1\left(\frac{3T}{8}\right) = A \cos\left(\frac{2\pi}{T} \cdot \frac{3T}{8}\right) = A \cos\left(\frac{3\pi}{4}\right) = A\left(-\frac{1}{\sqrt{2}}\right) = -\frac{A}{\sqrt{2}}$ $x_2\left(\frac{3T}{8}\right) = -A \sin\left(\frac{2\pi}{T} \cdot \frac{3T}{8}\right) = -A \sin\left(\frac{3\pi}{4}\right) = -A\left(\frac{1}{\sqrt{2}}\right) = -\frac{A}{\sqrt{2}}$ Indeed, their displacements are the same.

The final answer is $\frac{3T}{8}$.

Explanation of the solution:

1. Formulate SHM equations:
   • Body 1 starts at right extreme ($x=A$ at $t=0$), so its displacement is $x_1(t) = A \cos(\omega t)$.
   • Body 2 starts at equilibrium ($x=0$ at $t=0$) and moves left ($v<0$), so its displacement is $x_2(t) = -A \sin(\omega t)$.
2. Equate displacements: Set $x_1(t) = x_2(t)$. $A \cos(\omega t) = -A \sin(\omega t)$
3. Solve trigonometric equation: $\cos(\omega t) = -\sin(\omega t) \implies \tan(\omega t) = -1$.
4. Find smallest positive time: Let $\theta = \omega t$. The smallest positive solution for $\tan(\theta) = -1$ is $\theta = \frac{3\pi}{4}$. Substitute $\omega = \frac{2\pi}{T}$: $\frac{2\pi}{T} t = \frac{3\pi}{4}$ $t = \frac{3T}{8}$

Answer: The time when both the blocks will have the same displacement from the equilibrium position is $\frac{3T}{8}$.