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shaylasimonds6103
02.03.2020 •
Physics
A 3-kg block, attached to a spring, executes simple harmonic motion according to x= 2cos(50t) where x is in meters and t is in seconds. The spring constant of the spring is: (a) 1 N/m (b) 100 N/m (c) 150 N/m (d) 7500 N/m (e) none of the above
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Ответ:
The spring constant of the spring on which the block is attached is 7,500 N/m.
The given parameters;
mass of the block, m = 3 kgsimple harmonic equation, x = 2cos(50t)The angular frequency is calculated as;
ωt = 50t
ω = 50 rad/s
The spring constant of the spring is calculated as follows;
Thus, the spring constant of the spring is 7,500 N/m.
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Ответ:
Explanation:
w=50=√k/m=√k/3
SO
(50²)*3 is 7500N/m
Ответ:
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Hope I helped in some way :)