[3 marks] A particle of mass 100 g executes simple harmonic motion about x = 0 with angular frequency ω = 10 s-1. Its total mechanical energy is Etot = 0.45 J. Find the displacement of
the particle when its speed is 2 m/s. [3 marks]​

22.4 cm

Explanation:

The total mechanical energy of the particle Etot = K + U where K = kinetic energy = 1/2mv² where m = mass of particle = 100 g =0.1 kg and v = speed of particle = 2 m/s and U = potential energy = 1/2kx² where k = spring constant and x = displacement

Etot = K + U

U = Etot - K

U = Etot - 1/2mv²

Since Etot = 0.45 J, substituting the values of the other variables into the equation, we have

U = Etot - 1/2mv²

U = 0.45 J - 1/2 × 0.1 kg × (2 m/s)²

U = 0.45 J - 1/2 × 0.1 kg × 4 m²/s²

U = 0.45 J - 0.1 kg × 2 m²/s²

U = 0.45 J - 0.2 J

U = 0.25 J

Now U = 1/2kx² and the angular frequency ω = √(k/m) ⇒ ω²m = k

So, U = 1/2ω²mx²

x² = 2U/ω²m

x = (√2U/m)/ω

since ω = 10 rad/s, substituting the values of the variables into the equation, we have

x = (√2U/m)/ω

x = (√2 × 0.25 J/0.1 kg)/10 rad/s

x = (√0.5 J/0.1 kg)/10 rad/s

x = (√5 J/kg)/10 rad/s

x = 2.24/10 m

x = 0.224 m

x = 22.4 cm

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y