Question

A body of mass 1 kg is suspended from a spring of negligible mass. Another body of mass 500 g moving vertically upwards hits the suspended body with a velocity 3 ms^-1 and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision `10/pi` Hz, the amplitude of motion and the spring constant are respectively ____________.

Options

• 5 cm, 300 Nm^-1

• 10 cm, 300 Nm^-1

• 10 cm, 600 Nm^-1

• 5 cm, 600 Nm^-1

Answer 1

A body of mass 1 kg is suspended from a spring of negligible mass. Another body of mass 500 g moving vertically upwards hits the suspended body with a velocity 3 ms^-1 and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision `10/pi` Hz, the amplitude of motion and the spring constant are respectively 5 cm, 600 Nm^-1.

Explanation:

Let,

m = 1.0 kg and M = 1.0 kg+ 0.5 kg = 1.5 kg

Applying conservation of linear momentum,

mu = Mv

∴ (0.5)(3) = (1.5)v

∴ v = 1 m/s

As this value of v is maximum, we get,

`"v" = omega"A"`

`Rightarrow 1 = (20)"A"      .....(because "n" = 10/pi "Hz")`

`therefore "A" = 1/20  "m" = 5  "cm"`

`"As for a spring pendulum,"  omega = sqrt("k"/"M")`

`therefore "k" = "M" omega^2`

` = 1.5 xx (20)^2`

` = 600  "N"//"m"`