A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 18cm before coming to rest at its lowest point. It then continues to oscillate vertically.

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

2. Suppose the free-fall accelaration at some location on earth was exactly 9.8000 m/s2. What would it be at the top of a 1100-m-tall toawer at this location?

- Make a FBD of the hanging mass, There are two external forces acting on it that is the force of gravity due to its weight and the springs restoring force when its stretched to its lowest point. After hanging the mass on the spring a new equilibrium position is achieved which also causes the spring to stretch. We can apply the Equilibrium conditions at this point in vertical direction as:

k*x - m*g = 0

k = m*g / x

Where, x is the extension of the spring or mean stretch. which 0.5*amplitude (Lowest point). x = 9 cm

k = 0.23*9.81 / 0.09

k = 25.07 N/m

Answer: For part a we have the stiffness of the spring k = 25.07 N/m

- The amplitude of the oscillating motion is the half the amount of total stretch or the amount the spring extends above or below the mean position.

Amplitude = x = 9 cm

- The frequency of any oscillatory motion which can be modeled by SHM can be expressed as:

f = 1 / 2*p* sqrt ( k / m )

- Plug the values in:

f = 1 / 2*pi* sqrt (25.07 / 0.23 )

f = 1.66 Hz

Answer: For part c the frequency of oscillation is f = 1.66 Hz

Kelalove Kelalove · Answer 2 · 2021-02-10T18:29:05+01:00

Kelalove Kelalove

10-02-2021

Answer:

k = 25.07 N/m

Amplitude = 9 cm

f = 1.66 Hz

Explanation:

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