A 200 gram mass is attached to a horizontally aligned spring on a frictionless s

Question

A 200 gram mass is attached to a horizontally aligned spring on a frictionless surface. A force of 10 Newtons will stretch the spring 25 centimeters. If the spring is compressed to 15 centimeters and then released, calculate: the spring constant the angular frequency of the system (measured in radius per second) the frequency of the motion measured in Hertz the period of the motion the speed of the mass as it passes through the equilibrium point. the position of the mass one minute after the mass is released.

Explanation / Answer

Here ,

mass ,m = 200 gm = 0.2 Kg

for the spring ,

a)let the spring constant is k

k * x = F

k * 0.25 = 10

k = 40 N/m

the spring constant is 40 N/m

b)

Now , angular frequency = sqrt(k/m)

angular frequency = sqrt(40/.20)

angular frequency = 14.14 rad/s

c)

Frequency of motion = angular frequency/2pi

Frequency of motion = 14.14/(2 * pi)

Frequency of motion = 2.25 Hz

d)

period of motion = 1/frequency of motio

period of motion = 1/2.25 s

period of motion = 0.444 s

the period of motion is 0.444 s

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