A student decides to use a simple pendulum to check the shutter time of her camera. A meter stick is laid horizontally, and a simple pendulum is set up so that, when it is hanging vertically, the mass falls on the 50.0 cm mark of the meter stick. The pendulum is set into oscillation such that at its extreme positions, the mass falls on the 40.0 cm and 60.0 cm graduations. The student has chosen the length of the pendulum to be 99.3 cm.
She then takes a photography of the pendulum as it is oscillating. Her camera is set to a nominal 1/10 of second shutter time. By careful inspection of the photography, she observes that the mass moved directly from the 41.2 cm to 43.9 cm scale markings on the meter stick.
How long does it take for the mass to move directly from the 41.2 cm to the 43.9 cm scale markings?
omega = sq. root g / Length
y = 50 cm +10 cm(sin omega t)
The Attempt at a Solution
I received the 2nd equation from an example my prof did in class, and I followed his steps precisely, but received an answer that was way too high--a difference of 9.5 seconds. This makes the 1/10 second marking look ridiculous. I tried a few other things, including using a cosine function in case I chose the sine function incorrectly, and I just end up with a negative value for time. I promise that I followed my prof's instructions to the letter--does anyone have any better ideas? Because clearly, my answer doesn't make sense. (And yes, my calculator is in radian mode and I double-checked my decimal points and conversions between cm and m).