Particle in Simple Harmonic Motion

[Willjeezy]

Homework Statement

Not exactly sure why a time value of 0.500s is given, but I am positive it is why my answer isn't correct:

Q. a 1kg object is attached to a horizontal spring. The spring is initially stretched by 0.100m and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500s later. What is the maximum speed of the object?

Homework Equations

V_max = Aω
ω=√(k/m)
F_s=-kx

The Attempt at a Solution

Given:
t = 0.5s
Amplitude: 0.100m
m: 1kg

I can find k:
F_s=-kx
mg = -kx
(1kg * -9.8) = -k (0.100m)
k = 98 N/m

Find ω
ω=√(k/m)
= √(98/1)
= 9.90

V_max = Aω
=(0.1)(9.90)
=0.990 m/s.

However, the answer is 0.628m/s. I don't understand the purpose of "the next time the velocity is zero is 0.5s later" I interpret that as:

"it to 0.5seconds to reach negative amplitude", which to my understanding isn't needed?

 

[Willjeezy]

wait I see my problem, it is moving horizontally not vertically so mg = -kx doesn't apply here.
v = m / s
= 0.2m / 0.5s
= 0.4 m/s

don't know how to go about finding acceleration from here.

 

[voko]

Think. If the next time the speed is zero is in 0.5 second, then what is the frequency of the oscillation?

 

[Willjeezy]

to anyone wondering in the future...

the mass was actually the unneeded info:
ω=2∏f

frequency is = # cycles / s
= 1 hz in this situation

then use Vmax

SOLVED. Thanks.

 

[Nickie]

The time value of 0.500s is given to provide a specific point in time for the problem. In this case, it is the time at which the object's velocity is zero. This information is necessary to calculate the maximum speed of the object.

Your solution is correct in calculating the maximum speed, but the answer given (0.628m/s) may be taking into account the direction of the object's motion. Since the object is initially released from rest at the stretched position, it will first move towards the equilibrium point and then back towards the stretched position. Therefore, the maximum speed may be referring to the speed at the stretched position, which would be 0.628m/s. This is why the time value of 0.500s is important, as it allows us to calculate the speed at a specific point in time.

Additionally, the statement "the next time the speed of the object is zero is 0.500s later" means that the object's velocity will reach zero again after 0.500s from the initial release. This is because the object will oscillate back and forth between the stretched and equilibrium positions, with its velocity reaching zero at these points. This information is useful in understanding the motion of the object and can also be used to confirm the calculated maximum speed.

In summary, your solution is correct and the given time value is important for calculating the maximum speed at a specific point in time and understanding the motion of the object.