Question

A spring of unstretched length L and spring constant k is attached to a wall and an object of mass M resting on a frictionless surface (see figure). The object is pulled such that the spring is stretched a distance A then released. Let the +x-+x-direction be to the right.

What is the position function x()x⁡(t) for the object? Take x=0x=0 to be the position of the object when the spring is relaxed, and make the phase angle as simple as possible.
What is the velocity xvx of the object at =7/6 T, where TT is the period? Write the expression using fractions, not decimal values.
What is the acceleration xax of the object at =5/4 T?

Answer 1

The position function x(t) for the object is x(t) = A * cos(ωt), where A is the amplitude of the oscillation and ω is the angular frequency. The velocity function at t = 7/6 T is vx(t) = -Aωsin(7π/6), and the acceleration function at t = 5/4 T is ax(t) = -Aω²cos(5π/4).

Step-by-step explanation:

The position function x(t) for the object is given by the equation x(t) = A * cos(ωt + φ), where A is the amplitude of the oscillation, ω is the angular frequency (ω = 2π/T, where T is the period), t is the time, and φ is the phase angle. In this case, the phase angle can be simplified to zero since the object is released from rest at x = A. So the position function becomes x(t) = A * cos(ωt).

The velocity function vx(t) of the object at t = 7/6 T is given by the derivative of the position function with respect to time, which is vx(t) = -Aωsin(ωt). Plugging in t = 7/6 T, we get vx(t) = -Aωsin(7π/6).

The acceleration function ax(t) of the object at t = 5/4 T is given by the second derivative of the position function with respect to time, which is ax(t) = -Aω²cos(ωt). Plugging in t = 5/4 T, we get ax(t) = -Aω²cos(5π/4).

Answer 2

The position function x(t) for the object attached to the spring is x(t) = A * cos(ωt + φ). The velocity