Finding the Speed of a Mass on an Oscillating Spring

[shnav34]

A horizontal spring with stiffness 0.7 N/m has a relaxed length of 18 cm (0.18 m). A mass of 17 grams (0.017 kg) is attached and you stretch the spring to a total length of 21 cm (0.21 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 18 cm (0.18 m)?

I've attempted this problem several times, to no avail.

These are the equations I've been using.

Maximum speed = 0.5(mv^2) = 0.5(k)(x^2)
where k = spring constant and x is maximum stretch

V at any point = sqrt((k/m)*(xmax^2 - x^2))
k is spring constant, m is mass, x is equilibrium position and xmax is it's stretched distance

The last result I got was 0.0371 m/s. let me know if it's correct.

thanks in advance

 

[E=mc^84]

A horizontal spring with stiffness 0.7 N/m has a relaxed length of 18 cm (0.18 m). A mass of 17 grams (0.017 kg) is attached and you stretch the spring to a total length of 21 cm (0.21 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 18 cm (0.18 m)?

I've attempted this problem several times, to no avail.

These are the equations I've been using.

Maximum speed = 0.5(mv^2) = 0.5(k)(x^2)
where k = spring constant and x is maximum stretch

V at any point = sqrt((k/m)*(xmax^2 - x^2))
k is spring constant, m is mass, x is equilibrium position and xmax is it's stretched distance

The last result I got was 0.0371 m/s. let me know if it's correct.

thanks in advance

Hi, since the spring is being stretched from equilibrium you take the difference of the 2 values: x2-x1= (0.21-0.18)m = 0.03m and you should get 0.19m/s

 

[vela]

No, it's not correct. Show us what you actually did so we can see where the problem lies.

 

[E=mc^84]

I showed you, just plug in the value of x into the formula Kx^2=mv^2 and solve for v

 

[vela]

My earlier post wasn't in response to yours.

 

[E=mc^84]

My earlier post wasn't in response to yours.

oops, sorry:)

 

[shnav34]

i used the maximum speed equation as the speed would be greatest at x=0 (the relaxed spring length).

max speed
(mv^2) = k(x^2)

0.017 kg * v^2 = 0.7 N/m * 0.03m^2

from this I got the result of: 0.0371 m/s.
I haven't entered it to see, can someone confirm this or show a mistake

 

[E=mc^84]

i used the maximum speed equation as the speed would be greatest at x=0 (the relaxed spring length).

max speed
(mv^2) = k(x^2)

0.017 kg * v^2 = 0.7 N/m * 0.03m^2

from this I got the result of: 0.0371 m/s.
I haven't entered it to see, can someone confirm this or show a mistake

I think you are not squaring 0.03^2 = 0.0009

 

[vela]

i used the maximum speed equation as the speed would be greatest at x=0 (the relaxed spring length).

max speed
(mv^2) = k(x^2)

0.017 kg * v^2 = 0.7 N/m * 0.03m^2

from this I got the result of: 0.0371 m/s.
I haven't entered it to see, can someone confirm this or show a mistake

That's $v^2$.

 

[E=mc^84]

That's $v^2$.

No, the x needs to be squared on the other side of the equation

 

[vela]

Try what you think the OP did. You'll see you don't get the same answer. Then try solving for $v^2$ instead.