Answer
a) $\frac{1}{2}$
b) $\frac{1}{2}$
c) $\frac{1}{2}$
d) $\frac{1}{4}$
e) $1$
Work Step by Step
(a) We know that
$\omega=\frac{2\pi}{T}$
$\implies \omega^{\prime}=(\frac{2\pi }{2T})$
$\omega^{\prime}=\frac{1}{2}(\frac{2\pi}{T})$
$\omega^{\prime}=\frac{\omega}{2}$
Thus, the angular frequency changes by a factor of $\frac{1}{2}$.
(b) As $f=\frac{1}{T}$
$\implies f^{\prime}=\frac{1}{2T}$
$\implies f^{\prime}=\frac{1}{2}(\frac{1}{T})$
$\implies f^{\prime}=\frac{f}{2}$
The frequency changes by a factor of $\frac{1}{2}$.
(c) We know that
$v_{max}=A\omega$
from part(a), we can see that $\omega$ becomes half when $T$ is doubled.
$\implies v_{max}^{\prime}=A(\frac{\omega}{2})$
$\implies v_{max}^{\prime}=\frac{v_{max}}{2}$
The maximum speed changes by a factor of $\frac{1}{2}$.
(d) As $a_{max}=A\omega^2$
$\implies a_{max}^{\prime}=A(\frac{\omega}{2})^2$
$\implies a_{max}^{\prime}=A(\frac{\omega}{4})=\frac{1}{4}$
Thus, the maximum acceleration changes by a factor of $\frac{1}{4}$.
(e) We know that $E=\frac{1}{2}KA^2$. In this equation, the force is constant and the amplitude is constant; hence, the total energy remains the same even if the time period is doubled and we can say that the mechanical energy changes by a factor of 1.
