A small object is attached to the end of a string to form a simple pendulum. The period of its harmonic motion is measured for small angular displacements and three different lengths. For lengths of 1.000 m, 0.750 m, and 0.500 m, total time intervals for 50 oscillations are measured as 99.8 s, 86.6 s, and 71.1 s, respectively.

(b) Determine the mean value of [tex]g[/tex] obtained from these three independent measurements and compare it with the accepted value.

The mean value of g obtained from the measurements (9.852 m/s²) is very close to the accepted value of g (9.81 m/s²). The small difference may be due to experimental errors or uncertainties in the measurements.

To determine the mean value of g obtained from the three measurements, we can use the equation for the period of a simple pendulum, which is given by:

[tex]\[T = 2\pi\sqrt{\frac{L}{g}}\][/tex]

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Let's calculate the values of g for each measurement:

For the first measurement with a length of 1.000 m and a time interval of 99.8 s for 50 oscillations:

[tex]\[T = \frac{99.8 \, \text{s}}{50} = 1.996 \, \text{s}\][/tex]

Using the equation [tex]$T = 2\pi\sqrt{\frac{L}{g}}$[/tex], we can rearrange it to solve for g:

[tex]\[g = \frac{4\pi^2L}{T^2} = \frac{4\pi^2 \cdot 1.000 \, \text{m}}{(1.996 \, \text{s})^2} = 9.853 \, \text{m/s}^2\][/tex]

For the second measurement with a length of 0.750 m and a time interval of 86.6 s for 50 oscillations:

[tex]\[T = \frac{86.6 \, \text{s}}{50} = 1.732 \, \text{s}\][/tex]

Using the equation [tex]$T = 2\pi\sqrt{\frac{L}{g}}$[/tex], we can rearrange it to solve for g:

[tex]\[g = \frac{4\pi^2L}{T^2} = \frac{4\pi^2 \cdot 0.750 \, \text{m}}{(1.732 \, \text{s})^2} = 9.860 \, \text{m/s}^2\][/tex]

For the third measurement with a length of 0.500 m and a time interval of 71.1 s for 50 oscillations:

[tex]\[T = \frac{71.1 \, \text{s}}{50} = 1.422 \, \text{s}\][/tex]

Using the equation [tex]$T = 2\pi\sqrt{\frac{L}{g}}$[/tex], we can rearrange it to solve for g:

[tex]\[g = \frac{4\pi^2L}{T^2} = \frac{4\pi^2 \cdot 0.500 \, \text{m}}{(1.422 \, \text{s})^2} = 9.844 \, \text{m/s}^2\][/tex]

To find the mean value of g, we can simply calculate the average of the three values:

Mean value of

[tex]$g$ = $\frac{9.853 + 9.860 + 9.844}{3} = 9.852 \, \text{m/s}^2$[/tex]

Now, let's compare this mean value with the accepted value of g, which is approximately [tex]$9.81 \, \text{m/s}^2$[/tex].

Learn more about mean value

https://brainly.com/question/14882017

#SPJ11