Respuesta :
Answer: Option (B) is the correct answer.
Explanation:
Expression for centripetal acceleration is as follows.
a = [tex]r \omega^{2}[/tex] .......... (1)
Also, we know that
[tex]\omega = \frac{2 \pi}{T}[/tex] ........... (2)
Putting the value from equation (2) into equation (1) as follows.
a = [tex]r (\frac{2 \pi}{T})^{2}[/tex]
= [tex]r \frac{2 (\pi)^{2}}{4}[/tex]
As, [tex]a \propto \frac{1}{T^{2}}[/tex]
a = [tex]\frac{k}{T^{2}}[/tex]
or, [tex]aT^{2} = k[/tex]
Now, we will reduce a to [tex]\frac{a}{2}[/tex]. So, new value of [tex]T^{2}[/tex] will be equal to [tex]2T^{2}[/tex].
Therefore, value of new period will be as follows.
[tex]T' = \sqrt{2T^{2}}[/tex]
= [tex]\sqrt{2}T[/tex]
Thus, we can conclude that the new period is equal to [tex]T \sqrt{2}[/tex].
The new period should be [tex]T\sqrt{2}[/tex]
Calculation of new period:
The expression for centripetal acceleration should be [tex]a = r\omega^2[/tex]
Now
we know that [tex]w = 2\pi \div T[/tex]
Now here we put the values
[tex]a = r(2\pi\div T)^2\\\\= r (2(\pi)^2\div 4\\\\Since\ a \alpha \frac{1}{T^2}\\\\ aT^2 = K[/tex]
Now here we have to decrease to a by 2. So the new T value should be [tex]2T^2[/tex]
So, the new period should be
[tex]T = \sqrt{2T^2}\\\\ = \sqrt{2T}[/tex]
Learn more about the acceleration here: https://brainly.com/question/21799955
