Question

The amplitude of a particle executing shm is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 83cm/s,will be

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:

this is the answer

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