Question

The equation of motion for a simple harmonic oscillator (SHO) is:

Answer 1

Answer:

Explanation:

ω = $\sqrt{\frac{k}{m} }$

k = 2.5 N/m

m = 10 kg

$\omega = \sqrt{\frac{2.5}{10} }$

ω = .5 rad /s

x(t) = A cos(ωt + φ₀)

When t = 0 , x(t) = 0

0 = A cos(ωx 0 + φ₀)

cos φ₀ = 0

φ₀ = π /2

x(t) = A cos(ωt +π /2 )

Putting the value of ω

x(t) = A cos(.5 t +π /2 )

Differentiating on both sides

dx(t)/dt = - .5 A sin(.5 t +π /2 )

v(t) = - .5 A sin(.5 t +π /2 )

Given t =0 , v(t) = -5 m/s

-5 = - .5 A sin(.5 x0 +π /2 )

-5 = - .5 A sinπ /2

A = 10 m

x(t) = 10 cos( .5 t +π /2 )

b )

when t = π ( 3.14 s )

x(t) = -  10 m

when t = 2π ( 6.28s )

x(t) = 0

when t = 3π ( 9.42 s )

x(t) =  10 m

and so on