A trolley of mass 3.0 kg as shown in the fig. is connected two springs

1.	A trolley of mass 3.0 kg as shown in the fig. is connected two springs, each of spring constant 600 Nm-1. If the trolley is displaced from its equilibrium position by 5 cm and released, what is (a) the period of ensuing oscillations, and (b) the maximum speed of the trolley? How much energy is dissipated as heat by the time the trolley comes to rest due to damping forces?
Answer» Here the trolley is displaced slightly through and the other spring is compressed. Therefore. a net restoring force F comes into play such that F = -k_1y - k₂y = -(k₁ + k₂)y ⇒ ma = -(k₁ + k₂)y or, a = -({k₁ + k₁}/{m}) x y ...(i) Comparing equation (i) with the equation, a = -w²y, We get, ω = √{k₁ + k₁/m} Therefore, T = 2π/ω ⇒ T = 2π√{m/k₁ + k₂} Here, k₁ = k₂ = 600 Nm^-1, m = 3 kg (a) T = 2 x 3.14 x √{3/600 + 600} = 0.0314 s (b) When y = 5 cm = 5 x 10^-2 m The potential energy stored, U = 1/2 (k₁ + k₂)y₂ = 1/2 (600 + 600) x (5 x 10^-2)² = 1.5 J Now, maximum velocity is attained, when whole of the potential energy is converted into kinetic energy. i.e., 1/2 mv²_max = 1.5 Here, v_max = √{(1.5 x 2)/(m)} = √{(1.5 x 2)/(3)} = 1 ms^-1 (c) If the trolley comes to rest due to damping forces, the kinetic energy is converted into heat energy. Thus the heat energy produced = 1.5 J.

A trolley of mass 3.0 kg as shown in the fig. is connected two springs, each of spring constant 600 Nm-1. If the trolley is displaced from its equilibrium position by 5 cm and released, what is (a) the period of ensuing oscillations, and (b) the maximum speed of the trolley? How much energy is dissipated as heat by the time the trolley comes to rest due to damping forces?

Answer»

Here the trolley is displaced slightly through and the other spring is compressed. Therefore. a net restoring force F comes into play such that

F = -k_1y - k₂y

= -(k₁ + k₂)y

⇒ ma = -(k₁ + k₂)y

or, a = -({k₁ + k₁}/{m}) x y ...(i)

Comparing equation (i) with the equation,

a = -w²y,

We get, ω = √{k₁ + k₁/m}

Therefore, T = 2π/ω

⇒ T = 2π√{m/k₁ + k₂}

Here, k₁ = k₂ = 600 Nm^-1, m = 3 kg

(a) T = 2 x 3.14 x √{3/600 + 600} = 0.0314 s

(b) When y = 5 cm = 5 x 10^-2 m

The potential energy stored,

U = 1/2 (k₁ + k₂)y₂ = 1/2 (600 + 600) x (5 x 10^-2)²

= 1.5 J

Now, maximum velocity is attained, when whole of the potential energy is converted into kinetic energy.

i.e., 1/2 mv²_max = 1.5

Here, v_max = √{(1.5 x 2)/(m)}

= √{(1.5 x 2)/(3)} = 1 ms^-1

Thus the heat energy produced = 1.5 J.