c. A pendulum is pulled aside until it is 0.386 m above its lowest position and released. The pendulum is designed to emit sound waves at a frequency of 440 Hz; however, as it swings toward and away from an observer, the frequency appears to vary slightly. What would be the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer? Assume the speed of sound is 345 m/s.the pendulum is 0.250kg.

Question

c. A pendulum is pulled aside until it is 0.386 m above its lowest position and released.  The pendulum is designed to emit sound waves at a frequency of 440 Hz; however, as it swings toward and away from an observer, the frequency appears to vary slightly.  What would be the apparent frequency of the sound from the pendulum as it swings at its maximum speed toward an observer?  Assume the speed of sound is 345 m/s.the pendulum is 0.250kg.

anonymous · Answer

i know this equation is involved but i am confused fd=(Vw/Vw-Vs)fs

mrnood · Answer

You are given the height above datum at start. When it is released it will convert its PE into KE so you can calculate the speed when it is at lowest point(maximum speed)

The PE = mgh
KE = (mv^2)/2

so v=sqrt(2gh)

When you know the value of v you can put it in your formula for frequency.

anonymous · Answer

when you manipulate the formula why oes it become v=(2gh)^1/2 and not v=(2KE/m)^1/2 ? @mrnood

anonymous · Answer

never mind i got it!