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

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

magnets not strong enough

explanation: