# Tension in simple pendulum

## Homework Statement

A 100g pendulum bob hangs from a 1 meter string. What is the tension in the string when the pendulum is not moving (i.e. at the 'bottom')? What is the tension when it is at 30° to the vertical?

100g=.1kg
g=-9.8m/s^2
F_t=-mgcos(Θ)

## The Attempt at a Solution

Tension when not moving:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(0)=.98N

Tension at 30° to the vertical:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(30°)≈.15N

Is that correct?
My main dilemma is that my professor told us when a pendulum exceeds a 15° swing angle, that it can no longer be considered a simple pendulum and goes by more complicated physics, however, this pendulum swings to twice that angle.
Also, when I put this on a different site, someone answered and said "The tension at any angle cannot be known unless you also know its velocity at that point."

Thoughts?

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## Homework Statement

A 100g pendulum bob hangs from a 1 meter string. What is the tension in the string when the pendulum is not moving (i.e. at the 'bottom')? What is the tension when it is at 30° to the vertical?

100g=.1kg
g=-9.8m/s^2
F_t=-mgcos(Θ)

## The Attempt at a Solution

Tension when not moving:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(0)=.98N

Tension at 30° to the vertical:
F_t=-mgcos(Θ)=-(.1kg)(-9.8m/s^2)cos(30°)≈.15N

Is that correct?
My main dilemma is that my professor told us when a pendulum exceeds a 15° swing angle, that it can no longer be considered a simple pendulum and goes by more complicated physics, however, this pendulum swings to twice that angle.
Also, when I put this on a different site, someone answered and said "The tension at any angle cannot be known unless you also know its velocity at that point."

Thoughts?
Your professor is correct in terms of the simple pendulum approximation for period. In solving the differential equation to come up with T=2*Pi*sqrt(L/g), the assumption was made that sin(Θ)~~Θ, but really, that truncates the series representation of sin(Θ) at only one term. To be accurate, one should go out more terms, but it gets rather complicated quickly.

If the pendulum is swinging through the bottom, it is experiencing a greater tension because it also has a net acceleration (T exceeds mg, giving the centripetal acceleration up to the point of rotation), and by extension, it has a different tension at all points on the path of its swing if it has a non-zero velocity, changing velocity due to the centripetal acceleration.

The question would be clearer if it asked "What is the tension in the string when it is held stationary at 30deg from the vertical, by a force acting perpendicular to the vertical (i.e. your hand)." Otherwise, you have an unbalanced force and it makes for a bit of confusion.