Shm: A particle of mass 5 kg is suspended from a fixed point by a light elastic...

[markosheehan]

A particle of mass 5 kg is suspended from a fixed point by a light elastic stringwhich hangs vertically. The elastic constant of the string is 500 N/m.The mass is pulled down a vertical distance of 20 cm from the equilibriumposition and is then released from rest.(i) Show that the particle moves with simple harmonic motion.

what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)

 

[Ishuda]

A particle of mass 5 kg is suspended from a fixed point by a light elastic stringwhich hangs vertically. The elastic constant of the string is 500 N/m.The mass is pulled down a vertical distance of 20 cm from the equilibriumposition and is then released from rest.(i) Show that the particle moves with simple harmonic motion.

what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)

First, we need to assume that this experiment tales place in a zero gravity field and the 'massless' string supplies the only force to get the answer we want. So, what is the restoring force and how is it directed? Calling down positive and up negative, where does the weight end up going up after all of the strings force from being stretched is 'released' and at what time? What 'restoring force' has been accumulated during this part of the cycle. Then what happens, again and again and ...? What is the equation of the location of the weight as a function of time?

Hint: let a=acceleration, F=mass, m=mass, s=speed, and d=distance.
a = F/m
s = \(\displaystyle \int\, a\, dt\)
d = \(\displaystyle \int\, s\, dt\)

 

[Deleted member 4993]

A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m.The mass is pulled down a vertical distance of 20 cm from the equilibrium position and is then released from rest.(i) Show that the particle moves with simple harmonic motion.

what i tryed to do was find the force down and the force up find the resultant force and let it equal to F=5a and then that would prove it but to do this when i am finding the force in the string i need the natural length of the string but it is not given in the question. to find the force up i use F=k(length-natural length)

The original length of the string(spring) and gravitational acceleration will disappear from the ODE and you should find:

\(\displaystyle \displaystyle{\delta = \ \delta_o\ \ * \ \sin\left(\sqrt {\frac{k}{m}}\ t\right)}\)

where:

\(\displaystyle \delta_o \ \) = initial displacement due to weight

k = spring constant (= 500 N/m)