Tuesday

January 27, 2015

January 27, 2015

Posted by **Julia** on Thursday, January 24, 2013 at 9:24am.

If you instead pull the mass down 0.12 m from its equilibrium position and release it, what would the new period be?

- Physics -
**Joseph**, Thursday, January 24, 2013 at 11:34amI believe the period time will be tripled from the original time at 2.55s because the distance that the spring was puled down was 3 times the original length.

Answer= 2.55 seconds

- Physics -
**Elena**, Thursday, January 24, 2013 at 12:07pmThe period is

T= 2π•sqrt(m/k) =

=2•3.14•sqrt(0.22/12) =0.85 s.

It doesn’t depend on „x“

**Answer this Question**

**Related Questions**

physics - Suppose that a 200g mass (0.20kg) is oscillating at the end of a ...

Physics 203 - A block with mass of 5.0kg is suspended from an ideal spring ...

physics - So I'm doing a lab about simple harmonic information and I have to ...

Physics - So I'm doing a lab about simple harmonic information and I have to ...

physics - Consider an ideal spring that has an unstretched length l0 = 3.9 m. ...

Physics - Consider an ideal spring that has an unstretched length l0 = 3.9 m. ...

physics - Consider an ideal spring that has an unstretched length l0 = 3.9 m. ...

physics - A 1kg mass is suspended from a vertical spring with a spring constant ...

Physics - Consider an ideal spring that has an unstretched length l0 = 3.3 m. ...

physics - A mass of 2 kg is hung from the lower end of a vertical spring and ...