Question

A particle executes simple harmonic motion with an amplitude of 4 cm at the mean position the velocity of the particle is 10 centimetre per second the distance of the particle from the mean position when its speed become 5 cm per second is

Answer 1

Given,

Amplitude , A = 4cm

velocity of the particle at mean position , v₀ = 10cm/s

We know the relation between speed of particle , amplitude and distance of particle from mean position is given by

v = ω√{A² - x²}

Here ω is angular frequency,

speed at mean position , v₀ = ωA = 10 cm/s

speed at x distance from mean position , v = ω√{A² - x²} = 5 cm/s

So, ωA/ω√{A² - x²} = 10/5

⇒ A/√{A² - x²} = 2

⇒ 4/√(4² - x²) = 2

⇒ 4 = 4² - x²

⇒ x² = 12

⇒ x = ±2√3 cm

Hence, position of particle where speed = 5cm/s is ±2√3 cm

Hope this will help u.