Challenge Problem Physics?

Question

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure gravity. You have a long, thin wire labeled 1.15 g/m and a 1.22 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.22 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.22 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.

What value of gravity will you report back to headquarter

Answer 1

Since you're only vibrating half the string, and the dimensions of the "pulley" aren't specified, the pulley must be an "ideal" object touching the string only at exactly the midpoint of the string (else, the observation would be meaningless):

λ of 200 Hz = (the length of the string)/2

Answer 2

they could use the Newtonian equation: - s = u.t + 0.5.g.t^2 To degree gravity contained in the vacuum surrounding the Moon, all they'd ought to do is degree, very wisely, the time of fall from relax for a small weight. the burden will be dropped from an electronically opened seize - which in turn began an somewhat precise clock. The clock will be stopped, because the falling weight handed a basic beam, purely above the outdoors of the Moon. If the area 's', of fall for the burden, between the launch seize (clock initiate) and the basic beam (clock end) is wisely common then the equation purely will change into: - s = 0.5.g.t^2 the position 't' is the time of fall and 'g' is the Moon's gravitational acceleration (^2 ability squared). This equation rearranges to furnish: - g = 2.s/(t^2) therefore, both scholars ought to very purely degree the gravitational acceleration on the Moon.

Answer 3

JuST AsTRa is an a.s.s, and idi.ot, you can't use einstein's formula of energy for a simple harmonic question, just idio.tic.

Answer 4

E=MC2