Mass and spring on an inclined plane

In summary, the conversation discusses the derivation of the formula for simple harmonic motion, specifically the relationship between period, amplitude, and the parameters k and m. The formula is given as 1/T = 1/2pi*sqrt(k/m). The method for deriving this formula is through mathematical manipulation and substitution of a solution to the differential equation given in part b. The conversation also mentions the use of Latex for writing mathematical formulae.
  • #1
Kenny555
4
0

Homework Statement


[/B]
upload_2018-12-3_21-6-8.png

upload_2018-12-3_21-6-25.png


I got part A and B but in part c , but how to relate period to the equation in part b which is not similar to the cosine function of S.H.M . So i don't know how to obtain period from the equation

PS: In part b ,I use mgsin(theta)=-kx and then divided by m

Homework Equations

:
[/B]
Simple harmonic motion:
-amplitude*(omega)^2 cos(omega(t)+phase angle).

The Attempt at a Solution


[/B]
I directly use (2pi/T)=sqrt(k/m) to find period but i don't how to relate to the equation given in part b
 

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  • #2
Note: you can use the Σ sign on the menu to get access to symbols for omega, pi, etc. Also you can use Latex to set the formulae out neatly.

Q3. This is about the derivation of the formula ##\frac {1} {T} = \frac {1} {2\pi} \sqrt {\frac {k} {m} } ## that you mention.
You can look it up in your notes or on the web.
Essentially, you can show that ## x = A cos(ωt +φ) ## is a solution to the differential equation in part b. Then substituting this solution and equating its second derivative let's you get the T formula from comparing the expressions for the amplitude. (This is all mathematical jiggery pokery, which is why we just remember the resultant formula. It may be worth understanding, because SHM is common in other contexts, which all lead to the same sort of differential eqn.)These are the formulae you quote, written in Latex. If you copy them and put two hash signs in front and at the end, they'll show up properly and you can preview them.

\frac {1} {T} = \frac {1} {2\pi} \sqrt {\frac {k} {m} }

Aω^2cos(ω(t)+θ)
 
  • #3
Thanks a lot!
 
  • #4
Can u please tell me the solution of part B.
 
  • #5
Max Agarwal said:
Can u please tell me the solution of part B.
That's effectively a new question. Please post it as a new thread, filling in the template. In particular, be sure to post an attempt.
 

Related to Mass and spring on an inclined plane

What is the relationship between mass and spring on an inclined plane?

The relationship between mass and spring on an inclined plane is determined by the force of gravity, the angle of the incline, and the stiffness of the spring. As the mass increases, the force of gravity pulling the mass down the incline also increases. This in turn causes the spring to compress more, resulting in a greater restoring force.

How does the angle of the incline affect the behavior of the mass and spring system?

The angle of the incline plays a significant role in the behavior of the mass and spring system. As the incline angle increases, the force of gravity pulling the mass down the incline also increases. This causes the spring to compress more and results in a greater restoring force. The steeper the incline, the greater the restoring force and the faster the mass will move down the incline.

What is the role of gravity in a mass and spring system on an inclined plane?

Gravity is the driving force behind the behavior of a mass and spring system on an inclined plane. The force of gravity pulls the mass down the incline, causing the spring to compress and store potential energy. This potential energy is then converted into kinetic energy as the spring pushes the mass back up the incline.

How does the stiffness of the spring affect the motion of the mass on an inclined plane?

The stiffness of the spring, also known as its spring constant, affects the motion of the mass on an inclined plane by determining the amount of restoring force exerted on the mass. A stiffer spring will have a higher spring constant, resulting in a greater restoring force and faster motion of the mass down the incline.

What happens to the motion of the mass when the spring reaches its maximum compression on an inclined plane?

When the spring reaches its maximum compression, the restoring force will be at its maximum and the mass will reach its maximum velocity. This velocity will then decrease as the spring starts to expand and release the stored potential energy, eventually bringing the mass to a stop at the bottom of the incline. The process will then repeat as the spring oscillates back and forth.

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