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 = 6.9 seconds.

Answer 1

Answers:

a) 0.144 Hz

b) 0.904 rad/s

c) 11.818 m

d)$9.77 m/s^{2}$

Explanation:

The rest of the question is written below:

a) Calculate the frequency of oscillation (in Hertz) of the chandelier

b) Calculate the angular frequency $\omega$ of the chandelier in radians/ second

c) Determine the length $L$ in meters of the chandelier

d) That evening, while hanging out in JJ. 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 0.01 seconds slower, so the period is $T+0.01 s$. Determine the acceleration due to gravity in $m/s^{2}$ at the club.

a) The frequency $f$ has an inverse relation with the period $T$:

$f=\frac{1}(T}$ (1)

Where $T=6.9 s$

$f=\frac{1}(6.9s}=0.144 Hz$ (2)

b) The angular frequency $\omega$ is given by:

$\omega=2\pi f=\frac{2 \pi}{T}$ (3)

$\omega=2\pi (0.144 Hz)$ (4)

$\omega=0.904 rad/s$ (5)

c) Another expression for the period is:

$T=2 \pi \sqrt{\frac{L}{g}}$ (6)

Where:

$L$ is the length of the pendulum

$g=9.8 m/s^{2}$ is the mean acceleration due gravity

Isolating $L$:

$L=\frac{T^{2} g}{4 \pi^{2}}$ (7)

$L=\frac{(6.9)^{2} (9.8 m/s^{2})}{4 \pi^{2}}$ (8)

$L=11.818 m$ (9)

d) In this case the period of the pendulum is $T_{p}=T+0.01 s$. So, we will use equation (7) with this period and find $g$:

$g=\frac{4 \pi^{2}L}{(T+0.01 s)^{2}}$ (10)

$g=\frac{4 \pi^{2}(11.818 m)}{(6.9 s+0.01 s)^{2}}$ (11)

$g=9.77 m/s^{2}$ (12) This is the acceleration due gravity at the place, which is near the mean value of $9.8 m/s^{2}$