Solve Pendulum Problem: Tension 150 N, Mass .65 kg, 28m Apart

  • Thread starter Thread starter Kandycat
  • Start date Start date
  • Tags Tags
    Pendulum
AI Thread Summary
The problem involves a cord with a mass of 0.65 kg stretched between two supports 28 m apart, with a tension of 150 N. To determine the time it takes for a pulse to travel along the cord, the correct wave speed formula is v = √(F/(m/L)). Using this formula, the wave speed is calculated to be approximately 80.35 m/s. Consequently, the time for the pulse to travel the length of the cord is found to be 0.35 seconds. The initial confusion stemmed from incorrectly applying pendulum equations instead of focusing on the wave properties of the stretched cord.
Kandycat
Messages
21
Reaction score
0

Homework Statement


A cord of mass .65 kg is stretched between two supports 28 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other?


Homework Equations


umm... the only equation to find helpful was...

T = 2pi\sqrt{m/(mg/L)} = 2pi\sqrt{m/(F/L)}?


The Attempt at a Solution



Knowns:

F (tension) = 150 N
m = .65 kg
L = 28 m

2pi\sqrt{.65 kg/(150 N/28 m)} = 2.19 s?

Though the book says it is 35 s...

Help? Should there be another equation that I should be using or did I miss something?
 
Physics news on Phys.org
Hi Kandycat,

Kandycat said:

Homework Statement


A cord of mass .65 kg is stretched between two supports 28 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other?


Homework Equations


umm... the only equation to find helpful was...

T = 2pi\sqrt{m/(mg/L)} = 2pi\sqrt{m/(F/L)}?

This is not a pendulum, so these equations do not apply here. In this problem, a cord is stretched and a pulse is made to travel along the cord, so you want to find an expression for the wave speed of a stretched string. What would that be?
 
Oh... so is it v = \sqrt{F(t)/(m/L)}?

v = \sqrt{150 N/(.65 kg/ 28 m)} = 80.35 m/s

T = L/v = 28 m/80.35 m/s = .35 s

Thanks!

Sorry about that.
 
That looks right to me!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top