Help with right or wrong question in simple harmonic motion

In summary, the question is asking whether the periodic time for a spring pulled 5cm would be double the periodic time for a spring pulled 10cm, both with identical springs. The answer is wrong because the periodic time is not affected by the amplitude or force, as long as the spring constant is the same. This can be shown mathematically using the equation T=2pi*sqrt(m/k) for both springs. Therefore, the periodic times for both springs would be the same.
  • #1
madah12
326
1
I have this right or wrong question in physics that i know the answer to but not the way to solve it can you help me with it?:

2 identical springs were put Vertically. if one was pulled 5cm (spring1) and the other was pulled 10cm(spring2) and were left to move a simple harmonic motion would the periodic time for (spring2)be double the periodic time for (spring1).

The answer is wrong i think but i have to mathematically prove it so my teacher don't think i am doing it randomly
i think its wrong because according to the law for the periodic time
T=[tex]2\pi\ [/tex] x [tex]\frac{\sqrt{m}}{\sqrt{k}}[/tex]
and k =f/x and since x is in the root it even if we double it we won't get double T or something like that?
 
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  • #2
look ,the equation for the periodic time You wrote is correct.
It's stated that the springs are identical! means that the K is the same, K is constant for a spring, no matter how much you stretch it(edit-until you overstretch it,so hooke's law doesn't work).
although k=f*x and X is doubled, It only means that the f will be twice as less, the f is not constant! this is why the statement is wrong, They'll have the same periodic time
hope You understood,I'm kinda new to S.H.O.
good luck.
 
  • #3
so your telling me if we have 2 springs with the same constant and we pull one 1million meter and the other 1 centimeters they will have the same periodic time?
edit: oh and isn't hook's law f= -k * x
 
Last edited:
  • #4
Well,Lets stay in proportion.
You'll kill the spring if you pull it 2 much, but let's say you pull 1 spring 1 centimeter, and the other one 20 centimeters(and they made by the same manufacturer,by the same engineer and the same metal and the same bluh bluh bluh) yes, their periodic time will be the same,that's what I think, when you pull the spring more, it has more Elastic force( cause F=k*x and x is bigger now) so when it reaches the point where x is 1 centimeter(where the other spring started) it has a speed V and the force that acts in this position is like the forces acted on the other spring at first, but it has a speed, so it's not so "NOT OBVIOUS" when You think about it... but maybe I'm wrong(although I'm 99% tight this time).
Hope You understood.
 
  • #5
but in simple harmonic motion when the force increase so will the displacement right?
The restoring force must be proportional to the displacement so if the force increase so will the displacement and the amplitude right? so if we have bigger amplitude won't we need more time for one oscillation?
 
  • #6
madah12 said:
but in simple harmonic motion when the force increase so will the displacement right?
The restoring force must be proportional to the displacement so if the force increase so will the displacement and the amplitude right? so if we have bigger amplitude won't we need more time for one oscillation?

"so if we have bigger amplitude won't we need more time for one oscillation?"
I don't think so,if it would be like you stated, where is the A(amplitude) in the formula?
the amplitude is the most far point from the equilibrium if I'm not mistaken,It doesn't affect the T.
let's forget about springs,take a pendulum,It's periodic time is not dependent on the Amplitude..
 
  • #7
Dweirdo said:
"so if we have bigger amplitude won't we need more time for one oscillation?"
I don't think so,if it would be like you stated, where is the A(amplitude) in the formula?
the amplitude is the most far point from the equilibrium if I'm not mistaken,It doesn't affect the T.
let's forget about springs,take a pendulum,It's periodic time is not dependent on the Amplitude..

ok then assuming you are right how can i write all of that mathematically i mean the question also says if you were to put right or wrong you must give a mathematical reason
 
  • #8
madah12 said:
ok then assuming you are right how can i write all of that mathematically i mean the question also says if you were to put right or wrong you must give a mathematical reason

Assuming I'm right:
T1-T of spring one
T2-T of spring two
K=Both springs' constant.

T1=2pi *sqrt(m/k)
T2= 2pi*sqrt(m/k)
---------------------
T1/T2=1

:S
 

Related to Help with right or wrong question in simple harmonic motion

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where an object oscillates back and forth around an equilibrium point, following a specific pattern. It is characterized by a restoring force that is proportional to the displacement from the equilibrium point, causing the object to move in a sinusoidal motion.

2. How do you determine whether a question about simple harmonic motion is right or wrong?

In simple harmonic motion, the object's displacement, velocity, and acceleration can be described using trigonometric functions, such as sine and cosine. Therefore, a question about simple harmonic motion can be considered right if it involves these functions and the correct physical principles, such as Hooke's law. On the other hand, a question can be considered wrong if it involves incorrect or unrelated principles and equations.

3. What are some common misconceptions about simple harmonic motion?

One common misconception is that simple harmonic motion only occurs in pendulums. In reality, it can occur in various systems, such as springs, the Earth's tides, and even sound waves. Another misconception is that the amplitude of the motion is directly proportional to the frequency, when in fact, it is inversely proportional.

4. How do you find the period of a simple harmonic motion?

The period of simple harmonic motion is the time it takes for an object to complete one full cycle of its motion. It can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. Alternatively, if given the frequency (f) of the motion, the period can be calculated as T = 1/f.

5. Can simple harmonic motion occur without any external forces?

No, simple harmonic motion requires an external force to be present. This force is known as the restoring force and is responsible for pulling the object back towards the equilibrium point. Without this force, the object would not experience any oscillations and would simply remain at rest.

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