Question

Please help me with this problem

Part A As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.66 g/m and a 1.23 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 1.23 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 1.23 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 318 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work. What value of g will you report back to headquarters? Express your answer with the appropriate units. 1.157 NIR Submit Previous Answers Request Answer X Incorrect; Try Again; 7 attempts remaining < Return to Assignment Provide Feedback

Answer 1

given
mue = 1.66 g/m = 0.00166 kg/m

let L is the length of the string.

in second harmonic,

f = 2*v/(2*L)

f = 2*sqrt(T/mue)/(2*L)

f*L = sqrt(m*g/mue)

f^2*L^2 = m*g/mue

200^2*L^2 = 1.23*g/0.00166

L^2/g = 1.23/(0.00166*200^2

L^2/g = 0.018524   --------------(1)

Time period of oscillations, T = 318/200

= 1.59 s

we know, T = 2*pi*sqrt(L/g)

T^2 = 4*pi^2*L/g

==> L/g = T^2/(4*pi^2)

= 1.59/(4*pi^2)

= 0.040275

substitude above value in equation 1

L*(L/g) = 0.018524

L*0.040275 = 0.018524

==> L = 0.460 m

L/g = 0.40275

==> g = L/0.040275

= 0.46/0.040275

= 11.4 m/s^2 <<<<<<<<<<<<<<----------------------------Answer