Question

A 0.80 kg mass attached to an ideal spring oscillates horizontally with a period of 0.50 s and amplitude of 0.30 m. What is the spring constant of the spring?

Answer 1

Answer:

126.33 N/m

Explanation:

Angular frquency is given by

$f=\frac {2\pi}{t}$

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

$w=\frac {2\pi}{0.5}=4\pi rad/s\approx 12.57 rad/s$

Angular frequency is also given by

$w=\sqrt{\frac {k}{m}}$

Making k the subject then $k=mw^{2}$

Where k is spring constant, m is mass

Substituting m with 0.8 then

$k=0.8\times (4\pi)^{2}$

K=126.330936333944 N/m

Rounded off as 126.33 N/m