Question

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 1.73 g/m and a 1.30 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 1.30 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 1.30 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 g will you report back to headquarters?

Answer 1

The value of g is
`g =76.2 m/s^2`

Step-by-step explanation:

From the question we are told that

The mass of the weight is
`m = 1.30 kg`

The spring constant
`k = 1.73 g/m = 1.73 *10^(-3) \ kg/m`

The second harmonic frequency is
`f = 100 \ Hz`

The number of oscillation is
`N = 200`

The time taken is
`t = 315 \ s`

Generally the frequency is mathematically represented as

`f = (v)/(\lambda)`

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

`l = \lambda`

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is
`L`

So
`l = (L)/(2)`

The velocity of the vibrating string is mathematically represented as

`v = \sqrt{(T)/(\mu) }`

Where T is the tension on the string which can be mathematically represented as

`T = mg`

So

`v = \sqrt{(mg)/(k) }`

Then

`f = (v)/((L)/(2) )`

=>
`v = (fL )/(2)`

=>
`\sqrt{(mg)/(k) } = (fL)/(2)`

=>
`g = (f^2 L^2 \mu)/(4m)`

substituting values

`g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2`

`g = 3.326 m^(-1) s^(-2) L^2`

Generally the period of oscillation is mathematically represented as

`T_p = 2 \pi \sqrt{(L)/(g) }`

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

The period can be mathematically evaluated as

`T_p = (t)/(N)`

substituting values

`T_p = (315)/(200)`

`T_p = 1.575 \ s`

Therefore

`L = (1.575^2 * g )/(4 \pi ^2)`

`L = 0.0628 ^2 g`

so

`g = 3.326 m^(-1) s^(-2) L^2`

substituting for L

`g = 3.326 ((0.0628) g)^2`

=>
`g = (1)/((3.326)* (0.0628)^2)`

`g =76.2 m/s^2`