Question

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 5.82 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

Answer 1

We can use the equations for simple harmonic motion to solve this problem.

The equation for the velocity of an object in simple harmonic motion is given by:

v(t) = Vmax * sin(ωt)

where v(t) is the velocity at time t, Vmax is the maximum velocity, ω is the angular frequency, and t is the time.

The equation for the acceleration of an object in simple harmonic motion is given by:

a(t) = -ω^2 * x(t)

where a(t) is the acceleration at time t, x(t) is the displacement at time t, and ω is the angular frequency.

Since we know the maximum acceleration and maximum velocity, we can find the angular frequency using the fact that:

amax = ω^2 * xmax

where amax is the maximum acceleration, xmax is the maximum displacement, and ω is the angular frequency.

We can relate the maximum displacement and maximum velocity using the equation:

Vmax = ω * xmax

Solving for xmax, we get:

xmax = Vmax / ω

Substituting this into the equation for amax, we get:

amax = ω^2 * (Vmax / ω)

Solving for ω, we get:

ω = sqrt(amax / (Vmax / ω))

ω^2 = amax * ω / Vmax

ω^3 = amax * Vmax

ω = (amax * Vmax)^(1/3)

Plugging in the given values for amax and Vmax:

ω = (5.82 m/s^2 * 1.38 m/s)^(1/3)

ω ≈ 2.85 rad/s

Now we want to find the time elapsed between when the object's speed is at a maximum and the next instant when its acceleration is at a maximum.

When the speed is at a maximum, the velocity function is at its maximum or minimum value. This occurs when the derivative of the velocity, with respect to time, is zero. The derivative of v(t) = Vmax * sin(ωt) is:

dv/dt = ω * Vmax * cos(ωt)

Setting this equal to zero, we get:

ω * Vmax * cos(ωt) = 0

cos(ωt) = 0

t = (2n + 1) * pi / (2ω) , where n is an integer

Let's consider the first instant when the object's speed is at a maximum, which occurs when n = 0:

t1 = pi / (2ω) = pi / (2 * 2.85) ≈ 0.548 s

Now, when the acceleration is at a maximum, it is either at its maximum or minimum value which mean the displacement is zero (x(t) = 0). The displacement in simple harmonic motion is given by:

x(t) = xmax * cos(ωt)

Setting this equal to zero, we get:

xmax * cos(ωt) = 0

cos(ωt) = 0

t = (2n + 1) * pi / (2ω) , where n is an integer

Let's consider the first instant when the object's acceleration is at a maximum, which occurs when n = 1:

t2 = 3 * pi / (2ω) = 3 * pi / (2 * 2.85) ≈ 1.646 s

Now we can find the time elapsed between these two instants:

Δt = t2 - t1 = 1.646 s - 0.548 s ≈ 1.098 s

Therefore, the time elapsed between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum is approximately 1.098 seconds.