Find the spring constant and the amplitude of the simple harmonic moti

[Flinthill84]

Having a little trouble with this one:

A 4kg mass is attached to a spring and executes simple harmonic oscillation with a period of 1.50s. The total mechanical engery of the system is 12J. What is the spring constant. Determine the amplite.

I was able to find the amplitude which is 0.585m but am not able to find the spring constant. The formula that I have for spring constant you have to know the angular frequency so I do not think that is right. Can anyone help?

 

[Andrew Mason]

Having a little trouble with this one:

A 4kg mass is attached to a spring and executes simple harmonic oscillation with a period of 1.50s. The total mechanical engery of the system is 12J. What is the spring constant. Determine the amplite.

I was able to find the amplitude which is 0.585m but am not able to find the spring constant. The formula that I have for spring constant you have to know the angular frequency so I do not think that is right. Can anyone help?

Use the equation of shm: [itex]x = A sin \omega t [/itex] where [itex]\omega = \sqrt{k/m}[/itex]

 

[wais]

To find the spring constant in this situation, we can use the formula K = (4π²m)/T², where m is the mass and T is the period. In this case, m = 4kg and T = 1.50s. Plugging these values into the formula, we get K = (4π² * 4kg)/(1.50s)² = 16π²/2.25 = 22.4 N/m. So the spring constant in this system is 22.4 N/m.

To determine the amplitude, we can use the formula E = (1/2)K*A², where E is the total mechanical energy and A is the amplitude. Rearranging for A, we get A = √(2E/K). Plugging in the values for E and K that we just found, we get A = √(2*12J/22.4 N/m) = 0.585m. So the amplitude of the simple harmonic motion is 0.585m.

I hope this helps!