A 300 g object attached to a horizontal spring moves in simple harmonic motion with a period of 0.340 s. The total mechanical energy of the spring–object system is 3.00 J. (a) What is the spring constant (in N/m)?
(b) What is the amplitude (in m) of the motion?
(c) What is the percentage change in amplitude of motion if the total energy of the system is increased to 5.00 J?

Explanation:

angular velocity ω = 2π / T where T is time period of rotation .

= 2 x 3.14 / .34 = 18.47 rad /s

Total mechanical energy = 1/2 m ω²Α² where A is amplitude .

spring constant = k

m is mass

18.47² = k / .3

k = 102.34 N /m

b )

energy = 1/2 m ω²Α²

3 = .5 x 18.47² x A²

A² = .01759

A = .1326

13.26 cm .

c )

If energy becomes 5 J

5 = .5 x 18.47² x A²

A² = .0293

A = .1711

= 17.11 cm

percent increase

= (17.11 - 13.26 )x100 / 13.26

= 29%

98.4 m

explanation:

let s = distance up the plane  

then by conservation of energy  

1/2m(18.3)^2 = mg s sin(10)  

(18.3)^2 = 2g s sin(10)  

s = (18.3)^2/2gsin(10)  

= 98.39564526  

= 98.4 m