Recent content by warmfire540

so just leave it at: vmax=5.715(A)

Awesome! for part b, the max velocity is: well, i know two equations vmax=sqrt(k/m)A vmax=wA (w being angular velocity) however, i also know that velocity is greatest at the equilibrium position so how do i find this? do i need to know k? time out, i know k..k=25.48 so...

just like that?

okay, so k(-.3)=7.644 k=25.48 T=2pi*sqrt(m/k) (period of a loaded spring) T=2pi*sqrt(.78/25.48) T=1.099 f=1/T f=.9096

Oh, i see, well the weight acting down is .78*9.8=7.644 (mg) that means there must be an equal force acting up which must be 7.644 in the opposite direction. So is this equal force the spring constant? (k)

Ah i See! thank you soo much!

Huh? W=ma?

I still don't see.. I only know the weight, and the distance from it's equillibrium to rest position.

Yeah..so we sub v for the sqrt( T / mu ) so f2/f1= sqrt( T2 / mu )/sqrt( T1 / mu ) 1.0316=sqrt( T2 / mu )/sqrt( T1 / mu ) 1.0642=(T2/mu) / (T1/mu) 1.0642=T2/T1 So T1*1.0642 = T2 This is a 93.98% difference between strings, where as the difference in percent is 6.02%

I just don't get the k part..

Okay Well from here i see that f2/f1 = 392/380 = 1.0316 So the tuned string's tension is 1.0316 times more than the untuned strings tension. This is also seen as a 96.94%, or a 3.06% in tension.

How come? I don't get why we don't need lambda? What other equation is there? How do you do it "algebraically?" :cry:

Okay, well where would I derive a formula to find the tension, or therefore the change of tension in this case?

Yeah, but what is k? The spring constant? or does it have another meaning?

Yeah, you're right goodjob Just didn't see that while typing it out! Hope it helped though :-p