Respuesta :
The formula for the period of the pendulum is
[tex]T=2 \pi \sqrt{ \frac{L}{g} }= [/tex]
where L is the pendulum's length and g the gravitational acceleration.
Labeling with E the Earth and with M Mars, we can write the period of the pendulum on Earth as
[tex]T_E = 2 \pi \sqrt{ \frac{L}{g_E} } [/tex]
and the period of the pendulum on Mars as
[tex]T_M = 2 \pi \sqrt{ \frac{L}{g_M} } [/tex]
if we calculate the ratio, we get
[tex] \frac{T_M}{T_E}= \sqrt{ \frac{g_E}{g_M} } [/tex]
but we know that the gravitational acceleration on Mars is 0.37 times the gravitational acceleration on Earth:
[tex]g_M = 0.37 g_E[/tex]
Substituting into the formula, we find
[tex] \frac{T_M}{T_E}= \sqrt{ \frac{1}{0.37} }=1.64 [/tex]
And so, the period of the pendulum on Mars is
[tex]T_M = 1.64 T_E = 1.64 (1.67 s)=2.78 s[/tex]
[tex]T=2 \pi \sqrt{ \frac{L}{g} }= [/tex]
where L is the pendulum's length and g the gravitational acceleration.
Labeling with E the Earth and with M Mars, we can write the period of the pendulum on Earth as
[tex]T_E = 2 \pi \sqrt{ \frac{L}{g_E} } [/tex]
and the period of the pendulum on Mars as
[tex]T_M = 2 \pi \sqrt{ \frac{L}{g_M} } [/tex]
if we calculate the ratio, we get
[tex] \frac{T_M}{T_E}= \sqrt{ \frac{g_E}{g_M} } [/tex]
but we know that the gravitational acceleration on Mars is 0.37 times the gravitational acceleration on Earth:
[tex]g_M = 0.37 g_E[/tex]
Substituting into the formula, we find
[tex] \frac{T_M}{T_E}= \sqrt{ \frac{1}{0.37} }=1.64 [/tex]
And so, the period of the pendulum on Mars is
[tex]T_M = 1.64 T_E = 1.64 (1.67 s)=2.78 s[/tex]
The period of pendulum on Mars is about 2.78 s
[tex]\texttt{ }[/tex]
Further explanation
Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:
[tex]\boxed{E_p = \frac{1}{2}k x^2}[/tex]
where:
Ep = elastic potential energy ( J )
k = spring constant ( N/m )
x = spring extension ( compression ) ( m )
[tex]\texttt{ }[/tex]
[tex]\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}[/tex]
where:
T = period of simple pendulum ( s )
L = length of pendulum ( m )
g = gravitational acceleration ( m/s² )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
Period of Pendulum on Earth = T_E = 1.69 s
Acceleration of gravity on Earth = g_E = g
Acceleration of gravity on Mars = g_M = 0.37 g
Asked:
Period of Pendulum on Mars = T_M = ?
Solution:
[tex]T_E : T_M = 2\pi \sqrt{ \frac{L}{g_E} }} : 2\pi \sqrt{ \frac{L}{g_M} }}[/tex]
[tex]T_E : T_M = \sqrt{g_M} : \sqrt{g_E}[/tex]
[tex]1.69 : T_M = \sqrt{0.37g} : \sqrt{g}[/tex]
[tex]1.69 : T_M = \sqrt{0.37}[/tex]
[tex]T_M = 1.69 \div \sqrt{0.37}[/tex]
[tex]\boxed {T_M \approx 2.78 \texttt{ s}}[/tex]
[tex]\texttt{ }[/tex]
Learn more
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Young Modulus : https://brainly.com/question/9202964
- Simple Harmonic Motion : https://brainly.com/question/12069840
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Physics
Chapter: Elasticity
