A 0.45 mass is attached to a spring with a force constant of 26 and released from rest a distance of 3.2 from the equilibrium position of the spring.

Question

A 0.45 mass is attached to a spring with a force constant of 26 and released from rest a distance of 3.2 from the equilibrium position of the spring.

a.What is the maximum speed of the mass? (m/s)
b.How far is the mass from the equilibrium position when its speed is half the maximum speed? (cm)

Vmax=A sqrt(K/M) = .24m/s that's correct
half speed = .12m/s
A= vmax sqrt(m/K) = 1.6cm
but that's wrong?
CAn anyone tell me where i went wrong in the second part?
Thanks

Answer 1

I can only hope you units are standard units, kg, N/m, meters. Units are important.

b. If speed is half, KE is down to 1/4, which means PE is 3/4 max
or distance is 3/4 * 3.2 (meters?, centimeters?).

Answer 2

You need to get the right answer, 1.6 is not the correct answer.

Answer 3

To find the distance the mass is from the equilibrium position when its speed is half the maximum speed, you can use the equation x = Asin(ωt), where x is the displacement from the equilibrium position, A is the amplitude (maximum displacement), ω is the angular frequency, and t is the time.

First, let's find the angular frequency ω. It is given by ω = √(k/m), where k is the force constant and m is the mass.

ω = √(26/0.45) ≈ 10.16 rad/s

Next, let's find the time when the speed of the mass is half the maximum speed. Since the speed of the mass is directly proportional to the velocity v, which is given by v = ωAcos(ωt), we can set v = 0.5vmax and solve for t.

0.5vmax = ωAcos(ωt)

0.5(0.24 m/s) = 10.16 rad/s * A * cos(10.16 rad/s * t)

0.12 m/s = A * cos(10.16 rad/s * t)

Now, substitute the value for A as A = 0.032 m (1.6 cm).

0.12 m/s = (0.032 m) * cos(10.16 rad/s * t)

Now solve for t.

cos(10.16 rad/s * t) = 0.12 m/s / (0.032 m)

cos(10.16 rad/s * t) ≈ 3.75

Now, since the range of the cosine function is -1 to 1, the equation is not solvable. This means that the mass never reaches a speed that is exactly half the maximum speed.

So there is no distance from the equilibrium position when the speed is exactly half the maximum speed.

Therefore, there was no mistake in your calculation, and the answer is that there is no specific distance when the speed is halfway the maximum speed.

A 0.45 mass is attached to a spring with a force constant of 26 and released from rest a distance of 3.2 from the equilibrium position of the spring.

I can only hope you units are standard units, kg, N/m, meters. Units are important.

You need to get the right answer, 1.6 is not the correct answer.

peanus

To find the distance the mass is from the equilibrium position when its speed is half the maximum speed, you can use the equation x = Asin(ωt), where x is the displacement from the equilibrium position, A is the amplitude (maximum displacement), ω is the angular frequency, and t is the time.