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3 years ago

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A block with a mass of 0.400 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilib

rium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 25.0 cm from equilibrium is at t = 0.200 s.
(a) What is the block's period of oscillation?

(b) What is the the value of the spring constant?

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

a) T = 1,199 s b) k = 11.0 N / m

Explanation:

The oscillatory movement of a mass-spring system is described by the equation

x (t) = A cos (w t)

Where A is the amplitude of the movement. W is the angular velocity, which is given by

w = √ k / m

The amplitude of the system is the maximum elongation that the spring has before releasing it, in this case it is A = 50 cm = 0.50 m, the equation is

x = 0.50 cos (wt)

We can calculate the angular velocity with the point given by x = 0.250 m for t = 0.200s

w t = cos⁻¹ (x / A)

w = 1 / t cos⁻¹ (x / A)

w = 1/0.200 cos⁻¹(0.25 / 0.50)

Let's be careful because the angle is in radians

w = 5.24 rad / s

Angular velocity is

w = 2 π f = 2π / T

T = 2π / w

T = 2π / 5.24

T = 1,199 s

With the angular velocity equation we can take off the spring constant

w² = k / m

k = m w²

k = 0.400 5.24²

k = 11.0 N / m

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