Question

in the figure shown the spring constant is k the mass of upper disc is m and that of lower disc is 3 m the upper block is depressed down from its equilibrium position by a distance 5mg/k and released at t =0 find the velocity of m when normal reaction on 3m is mg

Answer 1

Given info : two discs of mass m and 3m are attached by a spring of spring constant k as shown in figure.the upper block is depressed down from its equilibrium position by a distance 5mg/k and released at t =0

To find : The velocity of m when normal reaction on 3m is mg.

solution : given normal reaction on 3m, N = mg

for disc of mass 3m,

at equilibrium,

N + spring force = weight

⇒mg + Kx = 3mg

⇒Kx = 2mg

⇒x = 2mg/K .....(1)

when m is in equilibrium,

normal reaction = spring force on m

⇒mg = Kx₀

⇒x₀ = mg/K

so total distance of 3m from the equilibrium position = δ + x₀ = 6mg/K

now from conservation of energy theorem,

1/2 K( δ + x₀)² - 1/2 Kx² = 1/2 mv² + mg(x + x₀ + δ)

⇒1/2 K(6mg/K)² - 1/2 K(2mg/K)² = 1/2 mv² + mg(8mg/K)

⇒1/2 K(32m²g²/K²) = 1/2 mv² + 8m²g²/K

⇒16m²g²/K = 1/2 mv² + 8m²g²/K

⇒8m²g²/K = 1/2 mv²

⇒v = 16mg²/K

⇒v = 4g√(m/K)

Therefore the velocity of m when normal reaction on 3m is mg, is 4g√(m/K)