Question

A 325 g object attached to a horizontal spring moves in simple harmonic motion with a period of 0.220 s. The total mechanical energy of the spring-mass system is 5.46 J. A. What is the maximum speed of the object (in m/s)?B) Find the force constant of the spring.C) Find the amplitude of the motion.

Answer 1

Step-by-step explanation:

Total mechanical energy = 1/2 m ω²A² where ω is angular frequency and A is amplitude .

Given

1/2 m ω²A² = 5.46

ω²A² = 2 x 5.6 / m

= 2 x 5.6 / .325

= 34.46

ωA = 5.87

maximum speed = ωA = 5.87 m /s

B )

angular frequency = 2π / T , T is period of oscillation .

= 2 x 3.14 / .22

= 28.54 s⁻¹

ω =
`\sqrt{(k)/(m) }`

k is force constant and m is mass

`28.54=\sqrt{(k)/(.325) }`

k = 264.72 N/m

C)

ωA = 5.87

28.54 X A = 5.87

A = .2056 m

= 20.56 cm .