A mass of 2.0kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (Take g=10m/s2)
Answers
Answer:
Explanation:
Mass = 2kg (Given)
Spring constant = 200N (Given)
Length of the spring = l (Given)
When m is placed over it, the equilibrium position becomes O' and if it is pressed from O' which is the equilibrium position to O", then O'O" is the amplitude. Thus,
OO' = mg/k
= 2×10200
= 0.10 m. mg = k x0.
If the restoring force mA ω² > mg, then the mass will move up with acceleration, detached from the pan.
i, e., A > g/ k/m
= A > 20/200 > 0.10 m.
The amplitude > 10 cm which means that the minimum is just greater than 10 cm. (The actual compression will include x0 also)
But in the case of amplitude, it is always measured from the equilibrium position with respect to which the mass is oscillating.
Answer:
the minimum amplitude is 10cm
Explanation:
Restoring force on spring is given by
F=ka
As restoring force is balanced by weight mg of block. For man to execute simple harmonic motion of amplitude A,
kA=mg⇒A= mg/k
Given, m=2kg, k=200N/m
and g=10ms^−2
∴ A=(2×10)/200
=10/100
=10cm
So, minimum amplitude of spring motion should 10 cm to detach the mass from the pan.
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