Answer
(a) $f = 1.54~Hz$
(b) $f = 2.08~Hz$
Work Step by Step
(a) We can find the force constant of the spring.
$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
$k = (m)(2\pi~f)^2$
$k = (0.750~kg)[(2\pi)(1.75~Hz)]^2$
$k = 90.7~N/m$
We can find the frequency when the mass is increased by 0.220 kg.
$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}\sqrt{\frac{90.7~N/m}{0.970~kg}}$
$f = 1.54~Hz$
(b) We can find the frequency when the mass is decreased by 0.220 kg.
$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}\sqrt{\frac{90.7~N/m}{0.530~kg}}$
$f = 2.08~Hz$
