Answer:
a) 0,1613 Hz
b) 1,01342 rad/sec
c) 9.5422 m
d) [tex]9.4314 m/sec^2[/tex]
Explanation:
In the Albert Michelson exhibit at Clark University, we know the period of oscillation of the chandelier is T = 6.2 seconds
The chandelier will be modeled as a simple pendulum
a) Since the frequency is the reciprocal of the period, we have
[tex]f=\frac{1}{T}=\frac{1}{6.2sec}=0,1613 Hz[/tex]
b) The angular frequency is computed as
[tex]w=2\pi f=2\pi (0,1613) = 1,01342\ rad/sec[/tex]
c) The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
Where L is its length and g is the local acceleration of gravity (assumed 9.8 [tex]m/sec^2[/tex] for this part)
We want to know the length of the pendulum, so we isolate L
[tex]L=\frac{T^2g}{4\pi^2}[/tex]
[tex]L=\frac{(6.2)^2(9.8))}{4\pi^2}=9.5422 m[/tex]
d) While hanging out in J.J. Thompson’s House O’ Blues, the new period is 6.2+0.11=6.32 sec. Since the chandelier is the very same (same length), we can assume the gravity is slightly different. We use again the formula for T, but now we'll isolate g as follows
[tex]g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2(9.5422m))}{(6.32sec)^2}[/tex]
Which results Â
[tex]g=9.4314\ m/sec^2[/tex]
The frequency of oscillation of the chandelier in Hertz will be 0.1613 Hertz
The frequency of oscillation of the chandelier will be calculated thus:
= 1/T = 1/6.2 = 0.1613 Hz
The angular frequency will be:
= 2πf = 2π × 0.1613 = 1.01342 rad/sec
The length of the simple pendulum will be:
= T²g/4π²
= (6.2)²(9.8) / 4π².
= 9.5422m
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