Answer:
Part a)
![F_(max) = 9.57 N](https://img.qammunity.org/2020/formulas/physics/high-school/quvqtjo841b25gpue9fh5ov07t9xkwicbi.png)
Part b)
![k = 172.4 N/m](https://img.qammunity.org/2020/formulas/physics/high-school/5ygzjo0xv6y7z9dszlqm04qljbori06vb4.png)
Step-by-step explanation:
Here we know that
Amplitude = 5.55 cm
Time period = 0.250 s
mass = 0.273 kg
Part a)
As we know that an object is executing SHM so the maximum acceleration of SHM is given as
![a_(max) = \omega^2 A](https://img.qammunity.org/2020/formulas/physics/middle-school/xvbvnns81j3rfzjj4s8izvi4t3qncx40qh.png)
![a_(max) = ((2\pi)/(0.250))^2(0.0555)](https://img.qammunity.org/2020/formulas/physics/high-school/fym8x1q9p2vqgmkrc3vuibrbs0texihh6u.png)
![a_(max) = 35 m/s^2](https://img.qammunity.org/2020/formulas/physics/high-school/5xrdnaeksoo58qm5mt94clspbpxcb6ksbc.png)
now we know that
![F_(max) = m a_(max)](https://img.qammunity.org/2020/formulas/physics/high-school/4ue7skfgo48mxwhdjigx41f3x6tp6hl7na.png)
here we have
![F_(max) = 0.273(35) = 9.57 N](https://img.qammunity.org/2020/formulas/physics/high-school/57dzj0u6937iuuxpl4gakwzilytwtu0j73.png)
Part b)
We know that angular frequency of SHM is given by following formula when it is a spring block system
![\omega^2 = (k)/(m)](https://img.qammunity.org/2020/formulas/physics/high-school/k55focw3mh8g37v4dhbsfiq6gdbb1fwchv.png)
so here we have
![((2\pi)/(T))^2 = (k)/(0.273)](https://img.qammunity.org/2020/formulas/physics/high-school/az10ujm1870y6x5xqfumr8f0yjc7j9r3ku.png)
![((2\pi)/(0.250))^2 = (k)/(0.273)](https://img.qammunity.org/2020/formulas/physics/high-school/ezl8xfmvacy5ocpoa5dahh4kneswc0kjq6.png)
![k = 172.4 N/m](https://img.qammunity.org/2020/formulas/physics/high-school/5ygzjo0xv6y7z9dszlqm04qljbori06vb4.png)