Question

A particle is executing shm with an amplitude of 4cm.at the mean position the velocity of the particle is 10 cm/sec.the distance of the particle from the mean position when its speed becomes 5cm/s 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

Answer 2

Answer: answer is 2√3 cm.

Explanation: