Question

A particle starts S.H.M. from the mean position. Its amplitude is $A$ and time period is $T$ . At the time when its speed is half of the maximum speed, its displacement $y$ is :
A. $\dfrac{A}{2}$
B. $\dfrac{A}{{\sqrt 2 }}$
C. $\dfrac{{A\sqrt 3 }}{2}$
D. $\dfrac{{2A}}{{\sqrt 3 }}$

Answer 1

To answer the question, we will first write the relationship between angular frequency and displacement, then differentiate the given equation with respect to time and find the maximum velocity, then find the equation for displacement when the speed is half of its maximum, and finally, by plugging in the parameters, we will arrive at our desired answer.

Complete step by step answer:
The following is the relationship between angular frequency and displacement:
$v = \omega \sqrt {{A^2} - {x^2}}$
Now, let us suppose that;
$x = Asin\omega t$
When we differentiate the given equation with respect to time, we get
$\dfrac{{dx}}{{dt}} = A\omega \cos \omega t$
Now, since $\dfrac{{dx}}{{dt}} = v = A\omega \cos \omega t$
According to the question, maximum velocity will be, when $\cos \omega t = 1$.......(Maximum possible value for $\cos \theta$ )
The maximum velocity value will be ${v_{\max }} = A\omega$.
When the speed is half of its maximum, the displacement is given as:
$v = \dfrac{{A\omega }}{2}$

Now squaring on the both side of the equation by putting value of $v$ from above
${A^2}{\omega ^2} = 4{\omega ^2}\left( {{A^2} - {x^2}} \right)$
Now, equating the equation
$\dfrac{{{A^2}{\omega ^2}}}{4} = {\omega ^2}({A^2} - {x^2})$
${\omega ^2}$ cancels out each other on both the side

\Rightarrow {x^2} = {A^2} - \dfrac{{{A^2}}}{4} \\\ $$ Taking lcm $$ \Rightarrow {x^2} = \dfrac{{4{A^2} - {A^2}}}{4} = \dfrac{{3{A^2}}}{4}$$ Therefore, from here the value of $x$ will be $$\therefore x = \dfrac{{A\sqrt 3 }}{2}$$ Therefore, its displacement $$y$$ is $$\dfrac{{A\sqrt 3 }}{2}$$. **Therefore,the correct option is C.** **Note:** It should be noted that one negative and one positive number will be used to solve for the displacement. However, we've just looked at the positive root. This is due to the fact that the options only mentioned positive displacements. In addition, the positive or negative sign of the displacement is determined by our decision.