A 0.36 kg object, attached to a spring with constant k=10n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude 0.082 m. What is it speed when it’s displacement is 0.041 m

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Explanation:

First, we need to determine the angular frequency of the system (), in radians per second:

(1)

Where:

- Spring constant, in newtons per meter.

- Mass, in kilograms.

If we know that and , then the angular frequency of the system is:

The kinematic formulas for the position (), in meters, velocity (), in meters per second, and acceleration of the object (), in meters per square second, are:

(2)

(3)

(4)

Where is the amplitude of the motion, in meters.

From (2) we determine the time associated with position (, ):

(5)

And the speed of the object is:

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

small = small current large mag large current

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s