Amass of 0.26 kg on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 m .a) find the velocity when it passes the equilibrium point. b) find the total energy of the system. c) find the spring constant. d) find the maximum acceleration of the mass.

a)V= 2.09 m/s

b)TE= 0.56 J

c)K=50.47 N/m

d)a=29.23 m/s²

Explanation:

Given that

m = 0.26 kg  ,  T= 0.45 s  ,A= 0.15 m

We know that time period given as

ω =Angular frequency

ω = 13.96 rad/s

The velocity at equilibrium

V=  ω A

V= 13.96 x 0.15

V= 2.09 m/s

The total energy TE

TE= 0.56 J

The spring constant K

Maximum stored energy in the spring

From energy balance

U= TE

K A² = m V²

K=50.47 N/m

The maximum acceleration a

a= ω² A

a = 13.96²  x 0.15 m/s²

a=29.23 m/s²

its c i just checked it

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr