Calculating Amplitude and Initial Values in Simple Harmonic Oscillator

In summary, the problem involves a simple harmonic oscillator with a mass of 2.60 kg and a spring constant of 380 N/m. At t = 2.30 s, the position and velocity of the block are x = 0.148 m and v = 4.080 m/s. Using the identity sin squared (a) + cos squared (a) = 1, where a is equal to (omega x t + phi), the amplitude of the oscillations can be found by solving for x subscript m. The frequency of the oscillator can also be found using the general form for the position of an oscillator.
  • #1
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A simple harmonic oscillator consists of a block of mass 2.60 kg attached to a spring of spring constant 380 N/m. When t = 2.30 s, the position and velocity of the block are x = 0.148 m and v = 4.080 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

For this problem, there was a hint online that for part A, said to use the identity sin squared (a) + cos squared (a) = 1, where a was equal to (omega x t + phi) (i'm not sure about the last greek symboL) and said to use this to relate the equation for velocity and position and solve for x subscript m. But I have no clue how to do this or any of the problem for that matter, but please help, I really need it
 
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  • #2
Can you find the frequency of the oscillator from the information given? The general form for the position of an oxcillator is

x = Asin(ωt + φ)

The first derivative wrt time gives the velocity. That will involve a cosine function. You do not know A or φ, but since you know the x and v at a particular time, you can find them. x and ω have different dimensions, so to use the hint you will want to use the squares of either x*ω and v, or x and v/ω.
 
  • #3
As OlderDan suggests:

The position of the oscillator contains a [tex]\sin(b)[/tex]
while the velocity contains a [tex]\cos(b)[/tex] where [tex]b = \omega t +\phi[/tex]
solve these two equations for the sin and the cos
square each of these two equations and and them up
the sum will be 1. This then gives you an equation for the amplitude of the motion.
 

Related to Calculating Amplitude and Initial Values in Simple Harmonic Oscillator

1. What is a simple harmonic oscillator?

A simple harmonic oscillator is a system that exhibits a periodic motion around an equilibrium point, where the restoring force is directly proportional to the displacement from that point. Examples include a mass attached to a spring or a pendulum.

2. How do you calculate the amplitude of a simple harmonic oscillator?

The amplitude of a simple harmonic oscillator can be calculated by finding the maximum displacement from the equilibrium point. This can be measured directly or calculated using the equation A = xmax - xeq, where xmax is the maximum displacement and xeq is the equilibrium point.

3. What is the equation for the displacement of a simple harmonic oscillator?

The displacement of a simple harmonic oscillator can be described by the equation x = A sin(ωt + φ), where x is the displacement, A is the amplitude, ω is the angular frequency, and φ is the phase angle. This equation shows that the displacement is a sinusoidal function of time.

4. How do you determine the initial values of a simple harmonic oscillator?

The initial values of a simple harmonic oscillator can be determined by analyzing the motion at a specific time. The initial displacement, velocity, and acceleration can be calculated using the equation x = A sin(ωt + φ), v = Aω cos(ωt + φ), and a = -Aω2 sin(ωt + φ), respectively. These initial values can also be measured directly from a graph of the motion.

5. How is the period of a simple harmonic oscillator related to its frequency?

The period of a simple harmonic oscillator is the time it takes for one complete oscillation, while the frequency is the number of oscillations per unit time. The relation between the two is T = 1/f, where T is the period and f is the frequency. This means that as the frequency increases, the period decreases and vice versa.

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