Answer
(a) $f(t)=5\cdot cos\sqrt {0.3}t $
(b) $f=\frac{1}{2\pi}\sqrt {\frac{k}{m}}$
(c) frequency will decrease, oscillation is slower.
(d) frequency will increase, oscillation is faster.
Work Step by Step
(a) Given $m=10g, k=3, a=5cm$, we can write the function as $f(t)=5\cdot cos\sqrt {3/10}t=5\cdot cos\sqrt {0.3}t $
(b) The period can be found as $p=\frac{2\pi}{\sqrt {k/m}}=2\pi\sqrt {\frac{m}{k}}$ and the frequency is the reciprocal of the period $f=\frac{1}{p}=\frac{1}{2\pi}\sqrt {\frac{k}{m}}$
(c) When the mass is increased, we expect the frequency to drop because $m$ is a denominator, which means that the oscillation is slower.
(d) When a spring with larger $k$ is used, the frequency will increase because $k$ is a numerator, which means that the oscillation is faster.
