# Finding Tension force in a SHM Pendulum

• JessicaHelena
In summary, a 110 kg panda is riding on a 3.0 m long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a 20° angle with the vertical. At that point, the tension in the ropes is approximately 1147.18 N.
JessicaHelena

## Homework Statement

A 110 kg panda is riding on a 3.0 m long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a 20° angle with the vertical. What the magnitude of the total tension in the ropes of the swing at that point?

m (mass of panda) = 110 kg
##\theta = 20 ^\circ##
L = 3.0 meters

## Homework Equations

Force diagrams (?) — I am not really sure if any specific equation is needed here.

## The Attempt at a Solution

Drawing a force diagram of the panda at that point, there is a tension force ##F_t## pointing towards the upper left along the rope, and a gravitational force ##F_g = mg## that points directly downwards. These forces do not balance each other out horizontally because of the horizontal component of ##F_t## but the ##F_{t(y)}## and ##F_g## do. Then we have ##F_t \cos \theta = F_g##. Rearranging, we have ##F_t = F_g / \cos \theta##. Plugging in the right values, ##F_t = (110)(9.8)/(\cos 20°)##. This gives approximately 1147.18 N.

For some reason, you divided by cosθ instead of multiplying.

@ Doc Al
Then is my derivation wrong? From $F_t \cos \theta = F_g$, to isolate F_t I think the only way is to divide by $\cos \theta$...

JessicaHelena said:
@ Doc Al
Then is my derivation wrong? From $F_t \cos \theta = F_g$, to isolate F_t I think the only way is to divide by $\cos \theta$...

Try this: Consider axes parallel and perpendicular to the rope. In particular, consider force components parallel to the rope. What must they add to? (What's the acceleration that direction?) Set up a force equation and you can solve for the tension.

Doc Al has told you the right way to analyse this, but I will point out what went wrong with your way:
JessicaHelena said:
##F_t \cos \theta = F_g##
If you are summing forces in the vertical direction then the resultant should correspond to any vertical acceleration. You seem to be assuming that is zero.

Re the LaTeX, in these forums you need to bracket it with a double hash (#) or double dollar. If you use the double dollar it will force a newline before and after.

## 1. How do you find the tension force in a SHM pendulum?

The tension force in a SHM (simple harmonic motion) pendulum can be found using the equation T = mgcosθ, where T is the tension force, m is the mass of the pendulum, g is the acceleration due to gravity, and θ is the angle of displacement from the equilibrium position.

## 2. What is SHM and how does it relate to a pendulum?

SHM (simple harmonic motion) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. This is similar to the motion of a pendulum, where the restoring force is the tension in the string and the displacement is the angle of the pendulum.

## 3. Can the tension force in a SHM pendulum be negative?

Yes, the tension force in a SHM pendulum can be negative if the angle of displacement is greater than 90 degrees. This indicates that the pendulum is moving in the opposite direction of the restoring force, which is caused by the force of gravity.

## 4. How does the length of a pendulum affect the tension force in a SHM?

The length of a pendulum does not directly affect the tension force in a SHM. However, it does affect the period of the pendulum's motion, which in turn affects the frequency and amplitude of the pendulum's oscillations.

## 5. Are there any real-life applications of finding tension force in a SHM pendulum?

Yes, the concept of SHM and finding tension force in a pendulum has many real-life applications, such as in clock mechanisms, seismographs, and musical instruments. Understanding the tension force in a pendulum can also help in designing and analyzing structures that are subjected to oscillatory motion.

• Introductory Physics Homework Help
Replies
20
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
234
• Introductory Physics Homework Help
Replies
9
Views
856
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
507
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K