Question

I need help with a physics question. mainly just part B and C

A block of mass ? = 2.00 kg is hung from a spring of stiffness ? = 1.5 × 10 N/m that hangs from the ceiling. The spring is massless and frictionless; also, ? ≈ 10.0 m/s.

a. (10 points) If the block is left to hang static in the new equilibrium position. Draw a free body diagram and find the extension of the spring.

b. (2.5 points) From the new position, the block is now pulled downwards a distance ? = 1.00 m and it is released from rest to oscillate. Show that the block will oscillate about the equilibrium position found in part a..

c. (2.5 points) Find the amplitude, angular frequency, frequency and period of oscillation of the block.

Will rate thanks for the help

Answer 1

a) Given ,

m = 2.00 kg = mass of the block

K = 1.5 × 10 N/m = force constant of the spring

Let , l = extension in the spring when mass m is suspended from it

When the spring is left to hang static ,

Magnitude of restoring force F = K×l (acts in the upward direction)

Weight mg acts in the downward direction

As the block is at rest , therefore

K×l = mg

l = mg / K

l = 2.00 × 10 / 1.5 × 10

l = 4/3 m

b) Initial Restoring force is given by

F = -K l

Now the block is further displaced through a distance A

New restoring force

F' = -K(l+A)

F' - F = -K(l+A) - (-Kl)

F' - F = -KA

As the resultant force F' - F is directly proportional to the negative of the displacemant A , therefore the block will oscillate about the equilibrium position and the amplitude of the oscillation would be A.

c) amplitude = maximum displacement through which the block is pulled down

amplitude of oscillation = 1 m

Angular frequency is given by

w = ( K/m )^1/2

w = (1.5×10/2)^1/2

w = 2.738 rad/s

w = 2.74 rad/s

Frequency is given by

= w/2

= 2.74/2

= 0.44 Hz

Time period of oscillation = 1/ frequency of oscillation

T = 1/

T = 2/w

T = 2×3.14/2.74

T = 2.29 s