When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. the mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. what is the period of the oscillations? 0.286 s

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π ×   ................2

put here value

period of oscillations = 2π ×  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

answergravity;