Question

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the block has a displacement of -0.70m, a velocity of -0.80 m/s, and an acceleration of +2.7 m/s2. What is the amplitude of the motion?

Answer 1

Step-by-step explanation:

Given

mass of block
`m=5.7\ kg`

at
`t=0 s`

displacement is
`x=-0.7\ m`

velocity
`v=-0.8\ m/s`

acceleration
`a=2.7\ m/s^2`

suppose
`x=A\cos (\omega t+\phi )` is the general equation of SHM

where A=amplitude

`\omega`=natural frequency of oscillation

therefore velocity and acceleration is given by

`v=-A\omega \sin (\omega t+\phi )`

`a=A\omega ^2\cos (\omega t+\phi )`

for t=0

`-0.7=A\cos (\phi )---1`

`v=-0.8=-A\omega \sin(\phi)---2`

`a=2.7=-A\omega ^2\cos(\phi )----3`

divide 1 and 3 we get

`\omega ^2=(27)/(7)`

`\omega =\sqrt{(27)/(7)}`

Now square and 1 and 2 we get

`(0.7)^2+((0.8)/(\omega ))^2=A^2`

`A^2=0.49+0.166`

`A=0.81\ m`