An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.1 rad/s. Calculate the distance d by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it.
To calculate the distance by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it, we need to use the equation for the displacement in simple harmonic motion.
The equation for the displacement of an object in simple harmonic motion can be written as:
x(t) = A * cos(ωt + φ)
Where:
x(t) is the displacement of the object at time t,
A is the amplitude (maximum displacement) of the motion,
ω is the angular frequency (2π times the frequency),
t is the time, and
φ is the phase constant.
In this case, the object is hanging stationary, which means it is at its equilibrium position where there is no displacement from the unstrained length of the spring. Therefore, we can consider this as the starting position or phase constant (φ) of 0.
Given that the angular frequency (ω) is 25.1 rad/s, and the object is hanging stationary, we can set up the equation as follows:
x(t) = A * cos(25.1t)
Since the object is hanging stationary, its displacement is 0, which means x(t) = 0.
0 = A * cos(25.1t)
To find the distance (d) by which the spring stretches from its unstrained length, we need to find the amplitude (A) in the equation.
The amplitude (A) is the maximum displacement from the equilibrium position, and in this case, it represents the distance the spring stretches when the object is allowed to hang stationary from it.
To find the amplitude (A), we can rearrange the equation and solve for it:
A = 0 / cos(25.1t)
Since cos(25.1t) can never be 0, it means that the amplitude (A) is 0. Therefore, the spring doesn't stretch from its unstrained length when the object is hanging stationary from it.
So, the distance (d) by which the spring stretches is 0.