a light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set vibrating and the plate is allowed to fall freely. 8 complete oscillations are counted when the plate falls through 10cm. What is the frequency of the fork ?
so the issue is how much time elapsed when it fell .1m.
d=1/2 g t^2
t=sqrt(2d/g)=sqrt(2*.1)/9.8) sec which is the time for 8 oscillations.
frequency=1/period=8/t
To find the frequency of the tuning fork, we can use the formula:
Frequency = Number of oscillations / Time
In this case, we are given the number of oscillations (8) and the distance the plate falls (10 cm). However, we do not have the time directly given.
To find the time, we can use the formula:
Distance = (1/2) * g * t^2
Where:
- Distance is the distance the plate falls (10 cm or 0.1 m),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time it takes for the plate to fall.
Rearranging the formula, we get:
t^2 = (2 * Distance) / g
Substituting the given values, we have:
t^2 = (2 * 0.1 m) / 9.8 m/s^2
t^2 ≈ 0.0204 s^2
Now we can find the time (t):
t = √(0.0204 s^2)
t ≈ 0.143 s
With the time (t) known, we can now calculate the frequency:
Frequency = Number of oscillations / Time
Frequency = 8 / 0.143 s
Frequency ≈ 55.944 Hz
Therefore, the frequency of the tuning fork is approximately 55.944 Hz.