Question

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 6.83 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum

Answer 1

t = 0.31s

Step-by-step explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

`v_(max)=\omega A\\\\a_(max)=\omega^2A`

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

`(v_(max))/(a_(max))=(\omega A)/(\omega^2 A)=(1)/(\omega)\\\\\omega=(a_(max))/(v_(max))=(6.83m/s^2)/(1.38m/s^2)=4.94(rad)/(s)`

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period by using the information about the angular frequency:

`T=(2\pi)/(\omega)=(2\pi)/(4.94rad/s)=1.26s`

Then the required time is:

`t=(T)/(4)=(1.26s)/(4)=0.31s`