college physics 1

Question

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m and undergoes simple harmonic motion with an amplitude of 12.5 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 32.0 cm/s.

(a) Calculate the mass of the block .

(b) Calculate the period of the motion.

(c) Calculate the maximum acceleration of the block.

Answer 1

Lisa,

This question requires us to use Newton's second law. With simple harmonic motion, the only force we have to worry about is the force of the spring. We can write

F = ma = -kx

F = ma is Newton's 2nd law, while -kx is the force of the spring (using Hooke's law).

Now, we know that a (acceleration) is the second derivative of x (the position).

m d²x = - kx

dt²

d²x = - (k/m)x (dividing by m on both sides)

dt²

The solution to the above differential equation is a cos/sin wave!

x(t) = Acos(ωt)

Let's take the derivative twice to see what happens

v(t) = -Aωsin(ωt)

a(t) = -Aω²cos(ωt) = -ω²x

Alright! Going back to our differential equation above, it looks like ω = √(k/m)!

Let's put all the pieces together. We are given k = 7 N/m. We also know that A = 12.5 cm. Finally, we are told when the block is halfway between the equilibrium and the end point (when x = A/2 = 6.25 cm), v = 32.0 cm/s.

Before we go any further, let's be VERY VERY careful with units. Notice that the spring constant has N/m (using m), but the other units use cm! Let's make sure everything is consistent before we go any further. Luckily, going from cm to m is not that hard. A = 12.5 cm is also 0.125 m, and v = 32 cm/s = 0.32 m/s.

First, at what time does the block reach x = 0.0625 m? (remember, let's keep all the units consistent!) Well, it's when

x(t) = 0.0625 m = 0.125 m cos(ωt)

0.5 = cos(ωt)

Or when ωt = 60 degrees (or π/3).

Plugging in π/3 into our velocity equation:

v(t) = -Aωsin(π/3)

Remember, we want speed (the absolute value of velocity), so we don't have to worry about the negative sign.

0.32 m/s = (0.125 m)ω √3 / 2

Let's isolate ω

(0.32 * 2)/(0.125 √3) = ω

5.12/√3 = ω (I like to keep the square root irrational terms around as long as possible, in case they go away!)

OK, and we know ω = √(k/m). So

5.12/√3 = √(k/m)

5.12²/3 = k/m (squaring both sides)

Don't forget we know k (7 N/m), so we can find m!

m = 7 * 3/5.12² = 21/26.2144 = 0.8011 kg

Great!!! The other two parts of the problem are very simple now that we've found m.

The period T = 1/f, where f is the frequency. We can get f from the angular frequency ω = 2πf.

ω = 2π/T

We saw above ω = 5.12/√3, so

T = 2π/ω = 2π / (5.12/√3) = 2.12 s

Finally, what is the maximum acceleration the block? It should just be the amplitude of a(t) = -Aω²cos(ωt), which is Aω².

.125m * (5.12/√3)², or 1.0923 m/s^2.

Answer 2

(a)

A = amplitude = 0.125 m

k = spring constant = 7.0 N/m

x = A/2 in part (a)

v = 0.32 m/s when x = A/2

Work with total energy to find m.

kA²/2 = mv²/2 + kx²/2 ⇒

m = (kA²/2 - kx²/2) / (v²/2)

m = k(A² - x²) / v²

You know k, A, x, and v. Now plug in the numbers and get m in kg.

(b)

T = 2π √(m/k)

Plug in the numbers and get T in seconds.

(c)

a_max = F_max / m = kA/m

Plug in the numbers and get a_max in m/s².