Question

In a certain oscillatory system, the amplitude of motion is $5m$ and the time period is $4s$ .The time taken by the particle for passing between points which are at distance of $4m$ and $2m$ from the center and on the same side of it will be
A. $0.30s$
B. $0.32s$
C. $0.33s$
D. $0.35s$

Answer 1

Using the distance amplitude formula we can solve this problem where we know the amplitude and the time period of the particle. Putting two distances one by one in place of distance in the formula we will get the respective time period of the particle. Now subtracting both the time periods we will get our solution.

Formula used:
Distance Amplitude Equation of an oscillatory particle:
$x = A\cos \left( {\omega t} \right)$
Where, Distance = $x$, Amplitude = $A$, Angular velocity of the particle = $\omega$ and Time period of the particle = $t$.

Complete step by step answer:
From the problem, there is an oscillatory system, the amplitude of motion is $5m$ and the time period is $4s$. Two different positions of the oscillatory particle are given.We need to calculate the time taken by the particle for passing between points which are at distance of $4m$ and $2m$ from the center and on the same side. Let $x_1$ and $x_2$ are the two different distances.
$x_1 = 4m$
$\Rightarrow x_2 = 2m$
We know the distance amplitude equation of an oscillatory particle as,
$x = A\cos \left( {\omega t} \right)$
Now we need to take two different cases for two different distance to find two different time period of the oscillatory particle,

Case I:
When the particle is at a distance of $4m$ then the time taken by the particle is,
$x_1 = A\cos \left( {\omega t_1} \right)$
We know the value of $A$ and $T$
$A = 5m$
$\omega = \dfrac{{2\pi }}{T}$
Where,
$T = 4s$
Now,
$\omega = \dfrac{{2\pi }}{{4s}}$
$\Rightarrow \omega = \dfrac{\pi }{2}{s^{ - 1}}$

Putting this in the distance equation we will get,
$4m = 5m\cos \left( {\dfrac{\pi }{2}{s^{ - 1}}t_1} \right)$
Rearranging the equation we will get,
$\dfrac{4}{5} = \cos \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_1} \right)$
Taking cos to other side we will get,
${\cos ^{ - 1}}\dfrac{4}{5} = \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_1} \right)$
$\Rightarrow 36.86 = \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_1} \right)$
$\Rightarrow 36.86 = 90t_1\,{s^{ - 1}}$
Rearranging the above equation we will get,
$\dfrac{{36.86}}{{90{s^{ - 1}}}} = t_1\,$
$\Rightarrow 0.41s = t_1\,$

Case II: When the particle is at a distance of $2m$ then the time taken by the particle is,
$x_2 = A\cos \left( {\omega t_2} \right)$
We know the value of $A$ and $T$
$A = 5m$
$\omega = \dfrac{{2\pi }}{T}$
Where, $T = 4s$
Now,
$\omega = \dfrac{{2\pi }}{{4s}}$
$\Rightarrow \omega = \dfrac{\pi }{2}{s^{ - 1}}$
Putting this in the distance equation we will get,
$2m = 5m\cos \left( {\dfrac{\pi }{2}{s^{ - 1}}t_2} \right)$

Rearranging the equation we will get,
$\dfrac{2}{5} = \cos \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_2} \right)$
Taking cos to other side we will get,
${\cos ^{ - 1}}\dfrac{2}{5} = \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_2} \right)$
$\Rightarrow 66.42 = \left( {\dfrac{{\pi {s^{ - 1}}}}{2}t_2} \right)$
$\Rightarrow 66.42 = 90t_2\,{s^{ - 1}}$
Rearranging the above equation we will get,
$\dfrac{{66.42}}{{90{s^{ - 1}}}} = t_2\,$
$\Rightarrow 0.74s = t_2$
Now, the time taken by the particle for passing between points which are at distance of $4m$ and $2m$ from the center and on the same side
$t = t_2 - t_1$
$\Rightarrow t = \left( {0.74 - 0.41} \right)s$
$\therefore t = 0.33\,s$

Therefore the correct option is $\left( C \right)$.

Note: Be careful while converting $\dfrac{\pi }{2}$ never change the $\pi$ into 3.14 always convert $\dfrac{\pi }{2}$ into its degree form because at the other side the value present in degree and we need to change the degree term so as to get the time in its SI unit that is second. If you write $\dfrac{\pi }{2}$ as $1.57$ then your solution will be increet and you will not get the time in its SI unit term.