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A block with a mass of 2 kg hangs withou...

A block with a mass of `2 kg` hangs without vibrating at the end of a spring of spring constant `500N//m`, which is attached to the ceiling of an elevator. The elevator is moving upwards with an acceleration `(g)/(3)`. At time `t = 0`, the acceleration suddenly ceases.
(a) What is the angular frequency of oscillation of the block after the acceleration ceases ?
(b) By what amount is the spring stretched during the time when the elevator is accelerating ?
(c )What is the amplitude of oscillation and initial phase angle observed by a rider in the elevator in the equation, `x = Asin (omega t + phi)` ? Take the upward direction to be positive. Take `g = 10.0 m//s^(2)`.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D

(a) Angular frequency
` omega = sqrt((k)/(m))` or `omega = sqrt((500)/(2))`
or `omega = 15.81 rad//s`
(b) Equation of motion of the block (while elevator is accelerating) is,

` kx - mg = ma = m(g)/(3)`
`:. x = (4mg)/(3k)`
`= ((4)(2)(10))/((3)(500)) = 0.053 m`
or `x = 5.3 cm`
(c) (i) In equilibrium when the elevator has zero acceleration, the equation of motion is,

`kx_(0) = mg`
or `x_(0) = (mg)/(k) = ((2)(10))/(500)`
`= 0.04m = 4cm`
`:.` Amplitude `A = x - x_(0) = 5.3 - 4.0`
` = 1.3cm`
(ii) At time `t = 0`, block is at `x = - A`. Therefore, substituting `x = - A` and `t = 0` in equation,
` x = Asin (omega t + phi)`
We get initial phase `phi = (3pi)/(2)`
.
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