Question

A 1.85 kg frictionless block is attached to an ideal spring with force constant 325 N/m . Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.5 m/s .Find (a) the amplitude of the motion, (b) the block’s maximum acceleration, and (c) the maximum force the spring exerts on the block.

Answer 1

Part(a): The amplitude of motion is 0.94 m.

Part(b): The maximum acceleration of the block is 165.13
`\bf{m~s^(-2)}`.

Part(c): The maximum force that the spring exerts on the block is 305.5 N.

Step-by-step explanation:

Part(a):

Given the mass (
`m`) of the block is 1.85 Kg, the force constant (
`k`) is 325
`N~m^(-1)`.Initially the spring is neither stretched nor compressed, which indicates that the block at this situation is in its equilibrium position where the maximum velocity of the block is
`v_(max) = 12.5~m~s^(-1)`. If '
`A`' be the amplitude of motion, then the velocity of the particle executing simple harmonic motion at any instant of position (
`x`) is

`v = \omega~{\sqrt{A^(2) - x^(2)}}......................................................(I)`

where
`\omega` is the natural angular frequency.

At equilibrium position, x = 0. So, the maximum velocity (
`v_(max)`), using equation (I) can be written as

`&& v_(max) = \omega~\sqrt{A^(2) - 0}\\&or,& v_(max) = \omega * A\\&or,& A = (v_(max))/(\omega) = v_(max) * \sqrt{(m)/(k)}\\&or,& A = (12.5)~m~s^(-1) * \sqrt{(1.85~Kg)/(325~N~m^(-1))} = 0.94~m`

Part(b):

The acceleration (
`a`) of a particle executing SHM is given by

`a = \omega^(2)~x = (k)/(m)~x................................................(II)`

The block will gain its maximum acceleration when it is at a distance equal to its amplitude. SO from equation (II), the maximum acceleration (
`a_(max)`) of the block is

`a_(max) = (k)/(m) * A = (325~n~m^(-1))/(1.85~Kg) * 0.94~m = 165.13~m~s^(-2)`

Part(c):

The block will experience a maximum restoring force (
`F_(max)`) when it is at a distance
`x = A`. So, the value of the maximum force is

`F_(max) = k * A = 325~N~m^(-1) * 0.94~m = 305.5~N`