A simple harmonic oscillator consists of a block of mass 4.40 kg attached to a spring of spring constant 110 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.126 m and v = 4.120 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

Question

A simple harmonic oscillator consists of a block of mass 4.40 kg attached to a spring of spring constant 110 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.126 m and v = 4.120 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

i got a to be:

(a) .833578

but i don't know how to find the rest
thanks!!!

Answer 1

To find the position and velocity of the block at t = 0 s, we can use the equations of motion for a simple harmonic oscillator.

The general equation for the position of a mass-spring system undergoing simple harmonic motion is given by:

x(t) = A * cos(ωt + φ)

where:
x(t) is the displacement of the block at time t,
A is the amplitude of the oscillations,
ω is the angular frequency,
t is the time, and
φ is the phase angle.

We can determine the amplitude A by using the information given:

At t = 1.00 s, the position x = 0.126 m and velocity v = 4.120 m/s.

Given that x = A * cos(ωt + φ), we have x(1) = A * cos(ω + φ) = 0.126 m.

We can also derive the equation for velocity as v(t) = -Aω * sin(ωt + φ):

v(1) = - Aω * sin(ω + φ) = 4.120 m/s.

We have two equations with two unknowns (A and ω), so we can solve for A.

First, we need to find ω using the relationship between angular frequency and spring constant:

ω = √(k / m)

where:
k is the spring constant, and
m is the mass of the block.

Plugging in the values, we have:

ω = √(110 N/m / 4.40 kg) = 5.00 rad/s (approximately)

Now we can solve for A using the equations:

0.126 m = A * cos(5.00 rad/s + φ)
4.120 m/s = - A * 5.00 rad/s * sin(5.00 rad/s + φ)

To solve for φ, we can divide the second equation by the first equation:

tan(5.00 rad/s + φ) = 4.120 m/s / (0.126 m * 5.00 rad/s)
tan(5.00 rad/s + φ) = 13.01

Using the inverse tangent function, we find:

φ = arctan(13.01) - 5.00 rad/s

Substituting the value of φ into the first equation, we can solve for A:

0.126 m = A * cos(5.00 rad/s + (arctan(13.01) - 5.00 rad/s))

Finally, we can calculate the position and velocity at t = 0 s using the amplitude A and the equations for position and velocity:

(b) position at t = 0 s:
x(0) = A * cos(ωt + φ)

(c) velocity at t = 0 s:
v(0) = -Aω * sin(ωt + φ)

With the obtained values of A, ω and φ, we can calculate the values for b and c.