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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

Respuesta :

PHG

Explanation:

w=50=√k/m=√k/3

SO

(50²)*3 is 7500N/m

onyebuchinnaji

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 kg
  • simple 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;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = \omega ^2 m\\\\k = (50)^2 (3)\\\\k = 7,500 \ N/m[/tex]

Thus, the spring constant of the spring is 7,500 N/m.

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