Answer
(a) $\frac{1}{\sqrt 2}$
(b) $5.90cm$
(c) $0.0702s$
Work Step by Step
(a) Using the equation:
$A=v_{\circ}\sqrt{\frac{m}{2K}}$
We see that if the spring force constant is doubled, then the amplitude will decrease by a factor of $\sqrt{2}.$
(b) We can find the new maximum compression of the spring as follows:
$A^{\prime}=v_{\circ}\sqrt{\frac{m}{2K}}$
$\implies A^{\prime}=\frac{1}{\sqrt 2}(v_{\circ}\sqrt{\frac{m}{K}})=\frac{A}{\sqrt 2}$
We plug in the known values to obtain:
$A^{\prime}=\frac{0.0835}{\sqrt 2}$
$A^{\prime}=5.90cm$
(c) We know that
$t=\frac{\pi}{2}\sqrt{\frac{m}{K}}$
We plug in the known values to obtain:
$t=\frac{\pi}{2}\sqrt{\frac{0.980}{490}}=0.0702s$
