motion in physics

Question

The position of a 0.34 kg object attached to a spring relative to the equilibrium position is given by [x = (0.49 m) cos(1.25πt)] a)Find the maximum displacement of the object relative to equilibrium in m. b) What is in meters/second the maximum speed of the oscillating object? Answer in meters/seconds. c) Find in meters/second the object’s speed at t = 0.01 s. d) What is in meters/seconds2 the maximum acceleration of the oscillating object? Provide your answers in 3 significant digits.

Answer 1

In general, an object attached to a spring has at time t position x(t)

governed by the following equation

x(t) = Acos(2πft), (1)

where A is the amplitude and f is the frequency.

In your case

A = 0.49 m, and

f = 0.625 s^-1

Using calculus you can derive that the velocity

v(t) = -2πfAsin(2πft) (2)

and acceleration

a(t) = -4π²f²Acos(2πft) (3)

a)

Maximum displacement is the amplitude in (1)

A = 0.49 m

b)

Maximum speed is the amplitude in (2)

2πfA = 2×π×0.625×0.49 ≈ 1.924 m/s

c)

v(0.01) = -2×π×0.625×0.49×sin(2×π×0.625×0.01) ≈ -0.0755 m/s

So the speed is 0.0755 m/s

d)

Maximum acceleration is the amplitude of (3)

4π²f²A = 4×π²×0.625²×0.49 ≈ 7.556 m/s²

Notice that the answers are independent of mass.