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3 years ago

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Suppose that you are on an unknown planet in a distant galaxy. Let the length of a physical pendulum be 1.02 m. In the physical

pendulum experiment, the period is determined to be 0.25 s.
What is the gravitational acceleration of this planet?

1 answer:

Natalka [10]3 years ago

6 0

Answer: acceleration due gravity = 619.5 m/s²

Explanation: The relationship between the period of a pendulum, length and acceleration due to gravity is given as

T =2π * (√l/g)

T = period = 0.25s

l = length of pendulum = 1.02m

g = acceleration due to gravity on the unknown planet

0.25 = 2π * (√1.02/g)

By squaring both sides, we have that

0.25² = 4π² * (1.02/g)

0.25² * g = 4π² * 1.02

g = 4π² * 1.02/ 0.25²

g = 40.26756/ 0.0625

g = 619.5m/s²

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a) The mass is 0.23 kg

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Explanation:

a)

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b)

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$v_{max}=1.25 m/s$ is the maximum speed

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The frequency in a spring-mass system is given by

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

where

k is the spring constant

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k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

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Substituting

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