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

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min