Oscillation of a spring - pulley system

[ineedhelp1234]

A massless spring of constant k=196 N/m is used to suspend the mass M=1.45 kg (including the pulley), as shown in the attached figure. What is the frequency of simple harmonic motion?


Relevant equations
f = 1/period
T = 2pi*sqrt(M/k)
Tension = 1/2Mg

What I tried...
T(tension on spring) = 1/2Mg <-- since each rope has equal tension?? with that, I found that the mass acting on the spring would be M/2 since the tension is half of the weight?? I feel like this may be where I'm going wrong..

f = 1/(period) = (1/2pi)*sqrt(k/m)
f = 2.62 /hz

I know I'm doing something wrong, I'm just not sure how what I need to change. Any explanations would be great!

 

[gneill]

The tension in the rope is going to vary as the mass moves up and down (this should be obvious, as the spring will change length and so must "administer" a force -kΔx accordingly).

Something to consider is the fact that, thanks to the rope and pulley, the other side of the rope (without the spring) is always going to have the same tension as the side with the spring. This assumes a massless pulley without friction or rotational inertia of course.

Since the tension versus displacement is the same for both ropes, the system behaves as though BOTH ropes have a spring in them! So what is the effective spring constant for two springs in parallel?

 

[ineedhelp1234]

Okay that makes a lot of sense :) Would the k of the spring then equal the k of the rope? So k(eff) = 2k ?

 

[gneill]

Okay that makes a lot of sense :) Would the k of the spring then equal the k of the rope? So k(eff) = 2k ?

Strictly speaking the ideal massless rope doesn't have a spring constant, but I understand what you mean Yes, the system behaves as though the "solid" rope were another spring of the same k.

In reality, that rope and pulley arrangement doubles the force that the mass feels from the spring for a given Δx of the spring.

 

[asteriatic]

Great job on using the relevant equations to try and solve for the frequency of the simple harmonic motion in the given system. However, there are a few errors in your approach. Firstly, the tension in the spring is not equal to half the weight of the mass. The tension in the spring is equal to the weight of the mass plus the weight of the pulley, which is equal to (M+M/2)g = 1.5Mg. This is because the mass of the pulley also contributes to the tension in the spring.

Secondly, when finding the period of the motion, you should use the total mass (M+M/2) instead of just the mass M. This is because the entire system is oscillating, not just the mass M. So the correct equation would be T = 2pi*sqrt((M+M/2)/k).

Finally, when finding the frequency, you should use the equation f = 1/T, which gives a frequency of approximately 2.62 Hz, as you correctly calculated.

In summary, the correct approach would be to use the total mass (M+M/2) and the total tension (1.5Mg) in the relevant equations to find the frequency of the simple harmonic motion in the given system. Keep up the good work!