# Tension in a string simple pendulum

## Homework Statement

Hi Guys
I obtained a slighty different answer. Can anyone kindly verify where I am going wrong? The question is stated thus:
A simple pendulum is of length 0.5m and the bob has mass 0.25kg. Find the greatest value for the tension in the string when the pendulum is set in oscillation by drawing the bob to one side through an angle of 5 degrees and releasing from rest. Explain where in the cycle the tension is greatest.

Te relationship for centripetal force involved is
T-mg = mv^2/r
Resolving the tension in the string into its components,
Tcos theta = mg
Tsin theta =mv^2/r
Dividing the above, we obtain
tan theta =v^2/gr
So, v^2 = gr tan theta
I substituted this value and other values for mass and g. I obtained 2.7N as my final answer. But it appears to be 2.5N. Any clarification will be appreciated. Thanks.

## The Attempt at a Solution

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Te relationship for centripetal force involved is
T-mg = mv^2/r
Resolving the tension in the string into its components,
Tcos theta = mg
Tsin theta =mv^2/r
Dividing the above, we obtain
tan theta =v^2/gr
So, v^2 = gr tan theta
I substituted this value and other values for mass and g. I obtained 2.7N as my final answer. But it appears to be 2.5N. Any clarification will be appreciated. Thanks.
When you say Tsin theta =mv^2/r, do you mean to imply that the horizontal component of tension is the centripital force?

When you say Tsin theta =mv^2/r, do you mean to imply that the horizontal component of tension is the centripital force?
Yes. That is my assumption. I have a feeling it may not be valid. Is tension equal to the weight at the point of rest?

Your are calculating the tension T at its highest position.
Maximum tension is at its lowest position.
At this position, the PE at the top is converted to maximum kinetic energy.
Find this velocity from conservation of energy.