If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced a distance 0.130 m from its equilibrium position and released with zero initial speed. Then after a time 0.803 s its displacement is found to be a distance 0.130 m on the opposite side, and it has passed the equilibrium position once during this interval.

A- Find the amplitude.
B- Find the period.
C- Find the frequency .

Please any help ?

Well, it seems like the object is having a bit of a bouncy ride, huh? Let's break it down and find some answers for you!

A) The amplitude of an oscillating object is the maximum displacement from its equilibrium position. In this case, since the object is displaced 0.130 m from its equilibrium position on both sides, the amplitude would be half of that distance. So the amplitude would be 0.130 m / 2 = 0.065 m.

B) The period of an oscillating object is the time it takes to complete one full cycle of oscillation. In this case, we are given that the object takes 0.803 s to go from one displacement of 0.130 m to the opposite displacement of 0.130 m. Since it passes the equilibrium position once during this interval, we know that it completes one full cycle in that time. Therefore, the period would be equal to 0.803 s.

C) The frequency of an oscillating object is the number of cycles it completes per second. We can calculate the frequency by taking the reciprocal of the period. So the frequency would be 1 / 0.803 s = 1.245 Hz (rounded to three decimal places).

So, in a nutshell:
A) The amplitude is 0.065 m.
B) The period is 0.803 s.
C) The frequency is 1.245 Hz.

Hope that helps! Remember, with a bouncy object like this, always keep an eye out for flying springs!

To solve this problem, we can use the equations of motion for simple harmonic oscillators.

A- Find the amplitude:
The amplitude can be found by using the displacement of the object at any point in time. In this case, we are given that the displacement is 0.130 m on the opposite side. The amplitude is always equal to the maximum displacement from the equilibrium position, so the amplitude is 0.130 m.

B- Find the period:
The period (T) is the time it takes for one complete oscillation. We are given that it passes the equilibrium position once during the interval of 0.803 s. Since the displacement is symmetrical about the equilibrium position, the oscillation from one side to the other takes half the time of one complete oscillation.
Therefore, the period is twice the given time interval: T = 2 * 0.803 s = 1.606 s.

C- Find the frequency:
The frequency (f) can be calculated using the formula:

f = 1 / T,

where T is the period.

Substituting the value of the period we found earlier:

f = 1 / 1.606 s = 0.622 Hz.

So, the frequency is 0.622 Hz.

Of course! I can help you solve this problem step by step.

Let's start with finding the amplitude of the oscillation.

Amplitude (A) is the maximum displacement of the object from its equilibrium position. In this case, we are given that the object is initially displaced a distance of 0.130 m from its equilibrium position.

So, the amplitude (A) is simply 0.130 m.

Next, let's calculate the period of the oscillation.

Period (T) is the time taken for one full cycle of the oscillation. We are given that after a time of 0.803 s, the object has completed one full cycle and reached a displacement of 0.130 m on the opposite side.

To calculate the period, we need to find the time taken for one complete cycle. Since the object has reached the opposite side after 0.803 s, we can assume that it takes 2 * 0.803 s, which is 1.606 s, to complete a full cycle.

So, the period (T) is 1.606 s.

Finally, let's find the frequency of the oscillation.

Frequency (f) is the number of oscillations per unit time. It is the reciprocal of the period.

To find the frequency, we use the formula f = 1/T. Substituting the value of T we found earlier, we have:

f = 1/1.606 s ≈ 0.623 Hz

So, the frequency (f) is approximately 0.623 Hz.

To recap:
A) The amplitude is 0.130 m.
B) The period is 1.606 s.
C) The frequency is approximately 0.623 Hz.