Question

A particle executes simple harmonic motion between $x=- A$ and $x=+ A$. If time taken by particle to go from $x=0$ to $\frac{A}{2}$ is 2 s; then time taken by particle in going from $x=\frac{A}{2}$ to $A$ is

• A. 1.5S
• B. 3s
• C. 4s
• D. 2s

Answer 1

The Correct Option is C

The correct answer is (C) : 4s

Let time from 0 to $A/2$ is $t_1$
$\&$ from $A/2$ to $A$ is $t_2$
then $\omega t_1=\pi /6$
$\omega t_2=\pi /3$
$\frac{t_1}{t_2}=\frac{1}{2}$
$t_2=2t_1=2\times 2=4\sec$

Answer 2

Let's consider a particle undergoing simple harmonic motion (SHM). Understanding the Problem: We are given that the time taken for the particle to travel from its equilibrium position (0) to half of its amplitude (A/2) is denoted as $t_1$. The time taken for the particle to travel from A/2 to the full amplitude (A) is denoted as $t_2$. Key Concepts: SHM Equation: The displacement of a particle in SHM is often described by the equation $x(t) = A \sin(\omega t)$, where:
$x(t)$ is the displacement at time $t$
$A$ is the amplitude
$\omega$ is the angular frequency Derivation: 1. Time to reach A/2:
When the particle is at A/2, we have:
$$\frac{A}{2} = A \sin(\omega t_1)$$
$$\frac{1}{2} = \sin(\omega t_1)$$
$$\omega t_1 = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$$
2. Time to reach A:
When the particle is at A, we have:
$$A = A \sin(\omega (t_1 + t_2))$$
$$1 = \sin(\omega (t_1 + t_2))$$
$$\omega (t_1 + t_2) = \sin^{-1}(1) = \frac{\pi}{2}$$
We also know that the time to reach A/2 from 0 is $\omega t_1 = \frac{\pi}{6}$.
The time to reach A from A/2 can be written as:
$$\omega t_2 = \omega (t_1 + t_2) - \omega t_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$
3. Ratio of Times:
We can find the ratio of $t_1$ to $t_2$:
$$\frac{\omega t_1}{\omega t_2} = \frac{\pi/6}{\pi/3} = \frac{1}{2}$$
$$\frac{t_1}{t_2} = \frac{1}{2}$$
4. Finding t2:
We are given that $t_1 = 2$ seconds.
Using the ratio, we can find $t_2$:
$$t_2 = 2 t_1 = 2 \times 2 = 4 \text{ seconds}$$ Conclusion: The time taken for the particle to travel from A/2 to A is 4 seconds.