# Equivalent length of a simple pendulum

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

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:

therefore, $$mg=2Tsin45$$

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

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:/)

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 $$d=l\sin\frac{\pi}{4}=\frac{l}{\sqrt{2}}$$.
• 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. – Michael Riberdy May 6 at 12:31