## Sounds of Music C

UTF-8 U+6211 U+662F
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### Re: Sounds of Music C

nobodynobody wrote:
October 15th, 2020, 5:29 am
Hello, sorry for being late!

Consider a massless pendulum with a length of 1.29 meters and a mass of 40.3 kg.

a) What is the period of the pendulum on earth (acceleration of 9.8)?
b) What is the period of the pendulum on the moon (acceleration of 1.62)?
c) If the string suspending the mass had a mass of 10.0kg, what is the new period(on earth)?
d) If the mass was removed (from question c), what is the new period of the pendulum, if it were on the moon?
Not really sure what this has to do with Sounds, but
a) $2\pi\sqrt{\frac{1.29}{9.8}} = 2.28\,\mathrm{s}$
b) $2\pi\sqrt{\frac{1.29}{1.62}} = 5.61\,\mathrm{s}$
c) 2.28 s
d) 5.61 s


nobodynobody
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### Re: Sounds of Music C

UTF-8 U+6211 U+662F wrote:
October 15th, 2020, 6:09 pm
nobodynobody wrote:
October 15th, 2020, 5:29 am
Hello, sorry for being late!

Consider a massless pendulum with a length of 1.29 meters and a mass of 40.3 kg.

a) What is the period of the pendulum on earth (acceleration of 9.8)?
b) What is the period of the pendulum on the moon (acceleration of 1.62)?
c) If the string suspending the mass had a mass of 10.0kg, what is the new period(on earth)?
d) If the mass was removed (from question c), what is the new period of the pendulum, if it were on the moon?
Not really sure what this has to do with Sounds, but
a) $2\pi\sqrt{\frac{1.29}{9.8}} = 2.28\,\mathrm{s}$
b) $2\pi\sqrt{\frac{1.29}{1.62}} = 5.61\,\mathrm{s}$
c) 2.28 s
d) 5.61 s

I have seen this type of stuff on tests (including 2019 nats) probably because it is simple harmonic motion. Your a and b are correct. C and D aren't because the period of the pendulum actually changes if the string has mass, I should specify that the string is uniform and rigid.
These users thanked the author nobodynobody for the post:
gz839918 (October 20th, 2020, 10:31 am)
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UTF-8 U+6211 U+662F
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### Re: Sounds of Music C

nobodynobody wrote:
October 16th, 2020, 5:43 am
UTF-8 U+6211 U+662F wrote:
October 15th, 2020, 6:09 pm
nobodynobody wrote:
October 15th, 2020, 5:29 am
Hello, sorry for being late!

Consider a massless pendulum with a length of 1.29 meters and a mass of 40.3 kg.

a) What is the period of the pendulum on earth (acceleration of 9.8)?
b) What is the period of the pendulum on the moon (acceleration of 1.62)?
c) If the string suspending the mass had a mass of 10.0kg, what is the new period(on earth)?
d) If the mass was removed (from question c), what is the new period of the pendulum, if it were on the moon?
Not really sure what this has to do with Sounds, but
a) $2\pi\sqrt{\frac{1.29}{9.8}} = 2.28\,\mathrm{s}$
b) $2\pi\sqrt{\frac{1.29}{1.62}} = 5.61\,\mathrm{s}$
c) 2.28 s
d) 5.61 s

I have seen this type of stuff on tests (including 2019 nats) probably because it is simple harmonic motion. Your a and b are correct. C and D aren't because the period of the pendulum actually changes if the string has mass, I should specify that the string is uniform and rigid.
Oh, the string has mass, I read that wrong
c) $2\pi\sqrt{\frac{\frac13(10.0)(1.29)^2+(40.3)(1.29)^2}{10.0\cdot9.8\cdot0.645+40.3\cdot9.8\cdot1.29}} = 2.24\,\mathrm{s}$
d) $2\pi\sqrt{\frac{\frac13(10.0)(1.29)^2}{10.0\cdot1.62\cdot0.645}} = 4.58\,\mathrm{s}$

Last edited by UTF-8 U+6211 U+662F on October 16th, 2020, 12:22 pm, edited 1 time in total.

nobodynobody
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### Re: Sounds of Music C

UTF-8 U+6211 U+662F wrote:
October 16th, 2020, 12:21 pm
nobodynobody wrote:
October 16th, 2020, 5:43 am
UTF-8 U+6211 U+662F wrote:
October 15th, 2020, 6:09 pm

Not really sure what this has to do with Sounds, but
a) $2\pi\sqrt{\frac{1.29}{9.8}} = 2.28\,\mathrm{s}$
b) $2\pi\sqrt{\frac{1.29}{1.62}} = 5.61\,\mathrm{s}$
c) 2.28 s
d) 5.61 s

I have seen this type of stuff on tests (including 2019 nats) probably because it is simple harmonic motion. Your a and b are correct. C and D aren't because the period of the pendulum actually changes if the string has mass, I should specify that the string is uniform and rigid.
Oh, the string has mass, I read that wrong
c) $2\pi\sqrt{\frac{\frac13(10.0)(1.29)^2+(40.3)(1.29)^2}{10.0\cdot9.8\cdot0.645+40.3\cdot9.8\cdot1.29}} = 2.24\,\mathrm{s}$
d) $2\pi\sqrt{\frac{\frac13(10.0)(1.29)^2}{10.0\cdot1.62\cdot0.645}} = 4.58\,\mathrm{s}$

Class of '23
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"No." - Albert Einstein

UTF-8 U+6211 U+662F
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### Re: Sounds of Music C

Define equal tempered. What is the difference between 12TET and 24TET (12-tone equal tempered and 24-tone equal tempered)?

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