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?

Given:

The angular frequency will be:

→

As we know,

→

hence,

The spring constant will be:

→

By substituting the values, we get

Thus the approach above is correct.      

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126.33 N/m

Explanation:

Angular frquency is given by

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

Angular frequency is also given by

Making k the subject then

Where k is spring constant, m is mass

Substituting m with 0.8 then

K=126.330936333944 N/m

Rounded off as 126.33 N/m

d.7.5cm is your answer