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leonid [27]
2 years ago
10

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?
Physics
1 answer:
sasho [114]2 years ago
8 0

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

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