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While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a simple pendulum) swings back and forth in the breeze once every
T = 6.2 seconds.Randomized VariablesT = 6.2 seconds(a) Calculate the frequency of oscillation (in Hertz) of the chandelier.(b) Calculate the angular frequency ω of the chandelier in radians/second.
(c) Determine the length L in meters of the chandelier. sig.gif?
(d) That evening, while hanging out in J.J. Thompson’s House O’ Blues, you notice that (coincidentally) there is a chandelier identical in every way to the one at the Michelson exhibit except this one swings back and forth 0.11 seconds slower, so the period is T + 0.11 seconds. Determine the acceleration due to gravity in m/s2 at the club.

Sagot :

Okpalawalter8idn

Answer:

a)  [tex]F=0.16Hz[/tex]

b)  [tex]\omega=1rad/s[/tex]

c)  [tex]L=9.54m[/tex]

d)  [tex]g=9.46m/s^2[/tex]

Step-by-step explanation:

From the question we are told that:

Time [tex]T =6.2[/tex]

a)

Generally the equation for Frequency F is mathematically given by:

 [tex]F=\frac{1}{T}[/tex]

 [tex]F=\frac{1}{6.2}[/tex]

 [tex]F=0.16Hz[/tex]

b)

Generally the equation for Frequency F is mathematically given by:

 [tex]\omega=2\pi f[/tex]

 [tex]\omega=2*314* 0.16[/tex]

 [tex]\omega=1rad/s[/tex]

c)

Generally the equation for Length of Chandelier L is mathematically given by:

Since

 [tex]T=2*3.142\sqrt{\frac{L}{g}}[/tex]

Therefore

 [tex]L=\frac{gT^2}{4 \pi^2}[/tex]

 [tex]L=\frac{9.8*(6.2)^2}{4 \pi^2}[/tex]

 [tex]L=9.54m[/tex]

d)

Since

Michelson exhibit except this one swings back and forth 0.11 seconds slower

Therefore

 [tex]T'=T+0.11[/tex]

 [tex]T'=6.31[/tex]

Generally the equation for acceleration due to gravity g is mathematically given by:

 [tex]g = \frac{4\pi ^2L}{( T')}^2[/tex]

 [tex]g = \frac{4\pi ^2L\*9.54}{( 6.31 s )}^2[/tex]

 [tex]g=9.46m/s^2[/tex]

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