# 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??\

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