# Solving for t in SHM (complex solution)

• mouser
In summary, the conversation was about finding the times when a function representing free undamped SHM equals zero. The poster was unsure of how to solve for t in this case and asked for advice. They were given a hint to rearrange the equation and eventually arrived at an incorrect solution. Upon realizing their mistake, they were able to correct it and find the correct solution.
mouser
Hey guys, this is my first post, I was hoping you all could offer some advice. I'm facing a problem involving free undamped SHM. Everything is working out well so far but I ran into a problem when trying to find t. Here's what I have:

x(t) = -2/3cos10t + 1/2sin10t

Now, if I want to find the times when x(t) = 0... how would I go about that? In a similar problem that only involved a single term, I was able to use cos^-1 to solve for t, but in this case would that still hold? Setting x(t) to 0 I get:

0 = -2/3cos10t + 1/2sin10t

But now I'm stuck.

Any help would be greatly appreciated!

Hi mouser! Welcome to PF!

mouser said:
0 = -2/3cos10t + 1/2sin10t

Hint: how might you re-arrange this equation?

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​

Thanks!

If I rearrange it to be:

2/3cos10t = 1/2sin10t

4/3cot10t = 1

3/4tan10t = 1

tan10t = 4/3

t = (tan^-1(4/3))/10

The answer is incorrect. Did I make a mistake in that algebra? Thank you for your help!

mouser said:
t = (tan^-1(4/3))/10

Looks ok to me!

Except you only have one solution … there should be infinitely many … what are the others?

(What were you actually asked for? Was it the times, or the period?)

The question was "at what time does the mass pass through the equilibrium position heading downward for the second time?"

I just figured out where I went wrong. The answer I got previously was only partially correct. For the correct answer I added the value for one full period and it checks out! eureka!

Thanks tiny-tim!

## 1. What is SHM and why is it important to solve for t?

SHM stands for Simple Harmonic Motion, which is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium. It is important to solve for t in SHM because it allows us to determine the time it takes for an object to complete one full cycle of motion, which is crucial for understanding and predicting the behavior of oscillating systems.

## 2. How do you solve for t in SHM with a complex solution?

In SHM, the equation for displacement is typically represented as x(t) = A*sin(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. To solve for t in a complex solution, you can use the inverse sine function and the quadratic formula to isolate t in the equation and find the time at which a specific displacement occurs.

## 3. What are the two possible solutions for t in SHM?

The two possible solutions for t in SHM are the simple solution (t = 2π/ω) and the complex solution (t = -φ/ω ± √(φ^2/ω^2 + A^2/ω^2)). The simple solution represents the time it takes for the object to complete one full cycle of motion, while the complex solution represents the time at which a specific displacement occurs.

## 4. Can you solve for t in SHM without knowing the amplitude or angular frequency?

No, in order to solve for t in SHM, you need to know both the amplitude and the angular frequency. These values are essential for determining the period of the motion and therefore, the time it takes for the object to complete one full cycle.

## 5. How does solving for t in SHM relate to real-world applications?

Solving for t in SHM has many real-world applications, such as predicting the motion of pendulums, springs, and other oscillating systems. It is also used in fields like engineering, physics, and astronomy to study and understand the behavior of various systems and phenomena. Additionally, the concept of SHM and solving for t can be applied to sound and light waves, which are essential in fields like acoustics and optics.

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