A block of mass m= 1.00kg is attached to the end of an ideal spring. Due to the

Question

A block of mass m= 1.00kg is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h= 9.00cm from its equilibrium length. (Figure 1) The spring has an unknown spring constant k. Take the acceleration due to gravity to be g = 9.81 m/s^2.Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency omega of the block's oscillations about its equilibrium position.

Explanation / Answer

A) Combine Hooke's Law and Newton's second law to obtain F = kx = mg (k)(0.09 m) = (1 kg)(9.81 m/s^2) k =109N/m B) The natural angular frequency of vibration for a spring-mass system is omega = SQRT(k/m) Thus, omega = 10.44 rad/s

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