Question

The time period of a particle in simple harmonic motion is 8s. At t=0, it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is:
$A.\quad \dfrac{1}{2}$
\[B.\quad \dfrac{1}{\sqrt{2}}\]
$C.\quad \dfrac{1}{\sqrt{2}-1}$
$D.\quad \dfrac{1}{\sqrt{3}}$

Answer 1

Hint: The problem is based on the concept of a particle executing Simple harmonic motion. The general formula for displacement of the particle in SHM is given by, $x=a\sin (\omega t+\phi )$. Finding out the frequency, which is the inverse of the time period and then the angular frequency will help in solving the problem.

Step by step solution:
Let’s start by taking the equation of motion of a particle executing simple harmonic motion, that is given by: $x=a\sin (\omega t+\phi )$. Here, x is the displacement with respect to the mean position, a is the amplitude, $\omega $is the angular frequency of the simple harmonic motion and $\phi $is the phase angle of the simple harmonic motion.
In the problem given, it states that the time period of the simple harmonic motion is 8 seconds. Hence, T=8s. Hence the frequency of the simple harmonic motion would be, $v=\dfrac{1}{T}\Rightarrow v=\dfrac{1}{8}Hz\Rightarrow v=0.125Hz$. Therefore the angular frequency of the simple harmonic motion would become, $\omega =2\pi v\Rightarrow \omega =2\pi (\dfrac{1}{8})\Rightarrow \omega =\dfrac{\pi }{4}Hz$.
The problem also states that, at t=0 seconds, the particle is at the mean position. Hence, x=0. Therefore, \[0=a\sin (\dfrac{\pi }{4}t+\phi )\Rightarrow \sin 0=a\sin (\dfrac{\pi }{4}(0)+\phi )\Rightarrow \sin 0=\sin \phi \Rightarrow \phi =0\].
Hence, the phase angle in this simple harmonic motion is zero.
The displacement at t=1s will be, \[{{x}_{1}}=a\sin (\omega (1))\Rightarrow {{x}_{1}}=a\sin \omega \Rightarrow {{x}_{1}}=a\sin \dfrac{\pi }{4}\Rightarrow {{x}_{1}}=\dfrac{a}{\sqrt{2}}\].
Similarly, the net displacement after t=2s will be, \[{{x}_{2}}=a\sin (\omega (2))\Rightarrow {{x}_{2}}=a\sin (2\omega )\Rightarrow {{x}_{2}}=a\sin (2\times \dfrac{\pi }{4})\Rightarrow {{x}_{2}}=a\sin (\dfrac{\pi }{2})\Rightarrow {{x}_{2}}=a\].
Hence, the displacement in the ${{2}^{nd}}$second is given by, $\Delta x={{x}_{2}}-{{x}_{1}}=a-\dfrac{a}{\sqrt{2}}=a(\dfrac{\sqrt{2}-1}{\sqrt{2}})$.
Therefore, the ratio of the distance travelled in the first second and the ${{2}^{nd}}$second is given by: $\dfrac{{{x}_{1}}}{\Delta x}=\dfrac{\dfrac{a}{\sqrt{2}}}{a(\dfrac{\sqrt{2}-1}{\sqrt{2}})}=\dfrac{1}{\sqrt{2}(\dfrac{\sqrt{2}-1}{\sqrt{2}})}=\dfrac{1}{\sqrt{2}-1}$. Hence, the Option C is correct.

Note: There are multiple different examples of particles executing simple harmonic motion in real life. The examples include simple pendulum, compound pendulum. However, in the real life scenarios, damped harmonic motion is observed. Only in a hypothetical scenario, the amplitude of simple harmonic motion remains constant over time.
Else, in a real life scenario, the amplitude keeps on damping until an external force is applied to keep the amplitude to remain constant.