Question

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.

Answer 1

Answer:

a)  $F=0.16Hz$

b)  $\omega=1rad/s$

c)  $L=9.54m$

d)  $g=9.46m/s^2$

Step-by-step explanation:

From the question we are told that:

Time $T =6.2$

a)

Generally the equation for Frequency F is mathematically given by:

 $F=\frac{1}{T}$

 $F=\frac{1}{6.2}$

 $F=0.16Hz$

b)

Generally the equation for Frequency F is mathematically given by:

 $\omega=2\pi f$

 $\omega=2*314* 0.16$

 $\omega=1rad/s$

c)

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

Since

 $T=2*3.142\sqrt{\frac{L}{g}}$

Therefore

 $L=\frac{gT^2}{4 \pi^2}$

 $L=\frac{9.8*(6.2)^2}{4 \pi^2}$

 $L=9.54m$

d)

Since

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

Therefore

 $T'=T+0.11$

 $T'=6.31$

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

 $g = \frac{4\pi ^2L}{( T')}^2$

 $g = \frac{4\pi ^2L\*9.54}{( 6.31 s )}^2$

 $g=9.46m/s^2$