A particle executes SHM with an amplitude of 4 cm.At the mean position

1.	A particle executes SHM with an amplitude of 4 cm.At the mean position, the velocity of the particle is10 cm/s. The distance of the particle from the meanposition when its speed becomes 5 cm/s is(a) v3 cm(c) 2(/3) crm(b) V5 cm(d) 2(V5) cm
Answer» 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

A particle executes SHM with an amplitude of 4 cm.At the mean position, the velocity of the particle is10 cm/s. The distance of the particle from the meanposition when its speed becomes 5 cm/s is(a) v3 cm(c) 2(/3) crm(b) V5 cm(d) 2(V5) cm

Answer»

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

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