Answer
(a) The fish weighs more than $4.90~N$
(b) The weight of the fish is $56.1~N$
Work Step by Step
(a) We can write an expression for the period:
$T = 2\pi~\sqrt{\frac{m}{k}} = 2\pi~\sqrt{\frac{W}{g~k}}$
We can see that increasing the weight will increase the period. Since the period is greater when the fish is on the spring, the fish must weigh more than $4.90~N$
(b) We can write an expression for the weight $W$ of the $4.90~N$ object:
$T_1 = 2\pi~\sqrt{\frac{W}{g~k}}$
$(\frac{T_1}{2\pi})^2 = \frac{W}{g~k}$
$W = \frac{T_1^2~gk}{(2\pi)^2}$
We can write an expression for the weight $W_f$ of the fish:
$T_2 = 2\pi~\sqrt{\frac{W_f}{g~k}}$
$(\frac{T_2}{2\pi})^2 = \frac{W_f}{g~k}$
$W_f = \frac{T_2^2~gk}{(2\pi)^2}$
We can find $W_f$:
$\frac{W_f}{W} = \frac{\frac{T_2^2~gk}{(2\pi)^2}}{\frac{T_1^2~gk}{(2\pi)^2}}$
$W_f = \frac{T_2^2}{T_1^2}~W$
$W_f = \frac{(11~s)^2}{(3.25~s)^2}~(4.90~N)$
$W_f = 56.1~N$
The weight of the fish is $56.1~N$
