Physics, 28.11.2019 19:31 Fireburntbudder

Asimple pendulum has a 100.0 g mass tied to it and 23.00 seconds are required to complete 10 oscillations.
(a) if g = 9.80 m/s^2, what is the length of the pendulum?
(b) when the mass is released from a certain hight, its speed as it passes through its lowest point is measured to be 1.620 m/s. calculate the tension in the string at the pendulum's lowest point,
(c) when the 100.00 g mass is removed from the string forming the pendulum, it is attached to a spring. if 23.00 seconds is required to complete 25 oscillations on the spring, calculate the spring constant,
(d) if you are to move pendulum to another planet where g = 6.58 m/s^2, how will period change?
(e) if you are to move spring mass system to another planet where g = 6.58 m/s^2, how will period change. show detailed calculations.

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Answer from: xxtonixwilsonxx

a) The length of the pendulum is 1.31 m

b) The tension in the string is 1.18 N

c) The spring constant is 4.66 N/m

d) The period of the pendulum increases from 2.30 s to 2.80 s

e) The period of the spring does not change

Explanation:

The frequency of oscillation of the pendulum is

$f=\frac{N}{t}=\frac{10}{23.0}=0.435 Hz$

where N = 10 is the number of oscillations in a time of t=23.0 s .

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum and g is the acceleration of gravity. Since the frequency is the reciprocal of the period,

$f=\frac{1}{T}=\frac{1}{2\pi} \sqrt{\frac{g}{L}}$

Since we know both f (the frequency) and g, we can solve the formula to find L, the length of the pendulum:

$L=\frac{g}{(2\pi f)^2}=\frac{9.80}{(2\pi (0.435))^2}=1.31 m$

The equation of motion for the pendulum when it passes through the lowest point is:

T-W = F_c

where

T is the tension in the string

W=mg is the weight of the pendulum, with m=100.0 g = 0.1 kg being the mass of the pendulum and g=9.80 m/s^2

$F_c = \frac{mv^2}{L}$ is the centripetal force, where

m = 0.1 kg is the mass of the pendulum

v = 1.620 m/s is the speed at the lowest point

L = 1.31 m is the radius of the circular trajectory, which is the length of the pendulum

Substituting and solving for T, we find

$T=mg+\frac{mv^2}{L}=(0.1)(9.80)+\frac{(0.1)(1.620)^2}{1.31}=1.18 N$

The frequency of oscillation of the spring is:

$f=\frac{N}{t}=\frac{25}{23.0}=1.087 Hz$

where N = 25 is the number of oscillations in a time of t=23.0 s .

The frequency of the spring can be written as

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m = 0.1 kg is the mass attached to the spring

Since we know f and m, we can re-arrange the equation to find k, the spring constant:

$k=(2\pi f)^2 m = (2\pi (1.087))^2(0.1)=4.66 N/m$

The period of the a pendulum, as we said earlier, is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

The length of this pendulum is

L = 1.31 m

On Earth, g=9.80 m/s^2 , so the period on Earth is

$T=2\pi \sqrt{\frac{1.31}{9.80}}=2.30 s$

Here the pendulum is moved to another planet, where the acceleration of gravity is

g=6.58 m/s^2

Substituting, we find the new period of the pendulum:

$T=2\pi \sqrt{\frac{1.31}{6.58}}=2.80 s$

So, the period increases from 2.30 s to 2.80 s.

The period of the spring is equal to the reciprocal of its frequency of oscillation:

$T=\frac{1}{f}$

We know that the frequency of oscillation is

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Therefore, we can write the period as

$T=2\pi \sqrt{\frac{m}{k}}$

We see that the period does not depend on the value of g, the acceleration of gravity: therefore, if we move the spring on another planet, the value of its period does not change.

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Asimple pendulum has a 100.0 g mass tied to it and 23.00 seconds are required to complete 10 oscilla...