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i was solving some question based on harmonic oscillations and a question popped up:

enter image description here

If the angle between the the wires and the surface is 45

and the mass of the bob is $m$ calculate the time period of the pendulum when displaced slightly in the horizontal direction(small oscillations)\

so, if u break the tension vectors acting on the bob as follows:

enter image description here

therefore, $mg=2Tsin45$

now if we see the pendulum from the left side such that only one wire faces u it looks planar and

enter image description here

since $theta$ is very small $sintheta= theta$ let the length of one string be $l$ $x/l=theta$ restoring force =Mgsin(theta)=mg*theta

ma=-mg*theta .......... T=2pi $(\frac{\sqrt l}{\sqrt g})$

but it turns out u need to use something called the effective length of a pendulum which is given by :(

L_effective=L/${\sqrt 2}$

can some one explain why am I supposed to find the effective length and what is wrong with my aproach??\

(please excuse my bad drawing, I had to draw with my mouse:/)

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You must consider the distance between the fulcrum and the oscillating mass. If both wires are of length $l$ and the angle between each wire and the horizontal is $\frac{\pi}{4}$ then the distance between the fulcrum and the oscillating mass is $d=l\sin\frac{\pi}{4}=\frac{l}{\sqrt{2}}$.

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  • $\begingroup$ why is that so? $\endgroup$ – Meet Lalwani May 6 at 10:38
  • $\begingroup$ Which part of my response are you asking about? $\endgroup$ – Michael Riberdy May 6 at 10:39
  • $\begingroup$ "You must consider the distance between the fulcrum and the oscillating mass" $\endgroup$ – Meet Lalwani May 6 at 12:12
  • $\begingroup$ "If both wires are of length l and the angle between each wire and the horizontal is π4 then the distance between the fulcrum and the oscillating mass is d=lsinπ4=l2√." $\endgroup$ – Meet Lalwani May 6 at 12:12
  • $\begingroup$ The normal length of a pendulum is the distance between the mass and the fulcrum. The question becomes: why would that change in this case? The second point is some trig based on your diagram. Drop a perpendicular from the vertex at the 90° angle to the hypotenuse. The length of that perpendicular is $lsin(45°)$ if $l$ is the length of the legs. That is the distance in question. $\endgroup$ – Michael Riberdy May 6 at 12:31

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