A hanging spring stretches by 30.0 cm when an object of mass 490 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 20.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later?

(b) Find the distance traveled by the vibrating object in part (a).
(c) Another hanging spring stretches by 30.5 cm when an object of mass 480 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 20.0 cm and released from rest to oscillate without friction. Find its position x at a time 84.4 s later.
(d) Find the distance traveled by the vibrating object in part (c).
(e) Why are the answers to the parts (a) and (c) different by such a large percentage when the data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close?

To solve this problem, we will use the concept of simple harmonic motion (SHM) and the equation for the displacement of an oscillating object.

(a) To find the position of the object at a moment 84.4 s later, we need to find its displacement from the equilibrium position.

Given that the spring stretches by 30.0 cm (or 0.3 m) when the object of mass 490 g (or 0.49 kg) is hung on it, we can use Hooke's Law to find the spring constant (k) of the spring.

Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement. Mathematically, it can be written as F = -kx, where F is the force exerted by the spring, x is the displacement from equilibrium position, and k is the spring constant.

In our case, when the object is at rest, the spring is stretched by 0.3 m. Therefore, we can write:

F = -k(0.3)

We also know that the weight of the object (mg) contributes to the force. So, we have:

-mg = -k(0.3)

Solving for k, we get:

k = mg / 0.3

Substituting the values, k = (0.49 kg)(9.8 m/s^2) / 0.3 = 16.06 N/m

Now, for simple harmonic motion, the equation for the displacement (x) of an oscillating object as a function of time (t) is given by:

x(t) = A * cos(ωt + φ)

where A is the amplitude, ω is the angular frequency, and φ is the phase constant.

Given that the object is pulled down an additional 20.0 cm (or 0.2 m) from the equilibrium position and released from rest, we can find the amplitude (A) as:

A = 0.2 m - (-0.3 m) = 0.5 m

Since the object is released from rest, its initial velocity (v₀) is zero. In SHM, the angular frequency (ω) is related to the spring constant (k) and the mass (m) of the object as:

ω = √(k / m)

Substituting the values, ω = √(16.06 N/m / 0.49 kg) = 1.99 rad/s

Now we can calculate the position (x) after 84.4 s by substituting the values into the equation:

x(84.4) = 0.5 cos(1.99 * 84.4)

Calculating this value will give us the answer to part (a).

(b) To find the distance traveled by the vibrating object in part (a), we can calculate the total distance covered during one complete oscillation.

Since the motion is harmonic, the object will complete one full oscillation after a time equal to the period (T). The period is given by:

T = (2π) / ω

Substituting the value of ω we found earlier, T = (2π) / 1.99 = 3.16 s

The distance traveled during one complete oscillation is equal to the circumference of the circular path followed by the vibrating object. The circumference is given by 2πA.

So, the distance traveled in one oscillation is 2πA = 2π * 0.5 = 3.14 m