Question

A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the block is sliding on a frinctionless surface; neglect gravity)
a) what is the period of the block's motion in seconds

b) if the blovk's max acceleration is 2 m/s^2, what is the amplitude of the motion (m)

Answer 1

Answer:

a) T=1.35s

b) amplitude = 0.0923m

Explanation:

m=300 gr

k=6.5 N/m

first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s   (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m