Question

An object is attached to a spring, the spring is stretched by 23.7 cm in the negative direction, and the object is released, oscillating in simple harmonic motion. After 0.317 s, it is again 23.7 cm from the equilibrium position, having passed through the equilibrium position once in those 0.317 s. Determine the speed of the object after 1.39 seconds. Number cm/s

Answer 1

Answer:83.17 cm/s

Step-by-step explanation:

Let positive x be negative direction and negative x be positive direction in this question

General equation of motion of SHM is

x=
`Asin\left ( \omega_(n)t\right )` --------1

where
`\omega`is natural frequency of motion given by

`\omega_n`=
`\sqrt{(k)/(m)}`

Where K is spring constant

here A=23.7cm

And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.

i.e. it covers this distance in
`(T)/(2)` sec

where T is the time period of oscillation i.e. returning to same place after T sec

therefore T=0.634 sec

differentiating equation 1 we get

v=
`A\omega_n`cos
`\left ( \omega_(n)t\right )`

and
`T* \omega_n`=
`2\pi`

`\omega_n`=9.911rad/s

`v`=
`23.7* 9.911cos\left (789.261\degree\right)`

v=83.17 cm/s