Simple Harmonic Motion and free-fall accelerations

In summary, the length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 m at Cambridge, England. The ratio of free-fall accelerations at these two locations can be found by using the formula g_C/g_T = l_C/l_T = 0.9942/0.9927 = 1 + 1.51 * 10^{-3}. Alternatively, the ratio can also be calculated using the formula T_C/T_T = 1, which results in the same conclusion.
  • #1
imnotsmart
40
0
A "seconds" pendulum is one that goes through its equilibrium position once each second. (The period of the pendulum is 2.000 s.) The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
Cambridge / Tokyo) = 1 +?

How do I work this problem? Any help would be appreciated.
 
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  • #2
As you observe, in Cambridge the value of g is higher than in Tokyo (g_C/g_T = 1 + ?).

Knowing that the periods are equal, then:

[tex]\frac{T_C}{T_T} = 1[/tex]

or


[tex]\frac{2\pi \sqrt{\frac{l_C}{g_C}}}{2\pi \sqrt{\frac{l_T}{g_T}}} = 1[/tex]

You can go on to find g_C/g_T in terms of the ratio l_C/l_T.
 
  • #3
Are you familiar with this formula?

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

[tex]T[/tex] is the peroid, so set it equal to two and solve for the value [tex]g[/tex] with your given [tex]L[/tex] value.
 
  • #4
theCandyman said:
Are you familiar with this formula?

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

[tex]T[/tex] is the peroid, so set it equal to two and solve for the value [tex]g[/tex] with your given [tex]L[/tex] value.

It is neater to express the ratio of free fall accelerations in terms of the ratio of the 1-sec pendulum lengths, than finding the free fall acceleration in each location and then computing the ratio.

Indeed the conclusion will be that:

[tex]\frac{g_C}{g_T} = \frac{l_C}{l_T} = \frac{0.9942}{0.9927} = 1 + 1.51 * 10^{-3}[/tex]
 
  • #5
Yes, yours is a much better way to compute it. I dislike dealing with ratios throught out a whole problem so I attempted it a different way.
 

Related to Simple Harmonic Motion and free-fall accelerations

1. What is Simple Harmonic Motion?

Simple Harmonic Motion is a type of periodic motion that occurs when a system is disturbed from its equilibrium position and experiences a restoring force that is proportional to the displacement from equilibrium. This results in a repetitive back-and-forth motion around the equilibrium point.

2. What factors affect the period of Simple Harmonic Motion?

The period of Simple Harmonic Motion is affected by the mass of the object, the strength of the restoring force, and the elasticity of the system. It is also independent of the amplitude of the motion.

3. How does Simple Harmonic Motion differ from free-fall acceleration?

Simple Harmonic Motion occurs when there is a restoring force acting on the system, while free-fall acceleration occurs when the only force acting on the object is gravity. In Simple Harmonic Motion, the object oscillates around an equilibrium point, while in free-fall acceleration, the object accelerates in a straight line towards the ground.

4. How does the period of Simple Harmonic Motion change with changes in the mass of the object?

The period of Simple Harmonic Motion is directly proportional to the square root of the mass of the object. This means that as the mass increases, the period also increases, and vice versa.

5. Can Simple Harmonic Motion and free-fall acceleration occur simultaneously?

No, Simple Harmonic Motion and free-fall acceleration cannot occur simultaneously. Simple Harmonic Motion requires a restoring force, while free-fall acceleration occurs when there is no other force acting on the object besides gravity. Additionally, Simple Harmonic Motion occurs in a confined system, while free-fall acceleration occurs in an unconfined system.

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