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.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west