A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t
1 answer:
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Answer:0.61 m
Explanation:
Given
Initial velocity (u)=7.40 m/s
launch angle
height of fence=0.7 m
Trajectory of Projectile is given by
For x=2.1 m
At x=2.1 m Y=1.31
therefore Fox clear fence by a margin =1.31-0.7=0.61 m
For a cylinder that has both ends open resonant frequency is given by the following formula:
Where n is the resonance node, v is the speed of sound in air and L is the length of a cylinder.
The fundamental frequency is simply the lowest resonant frequency.
We find it by plugging in n=1:
To find what harmonic has to be excited so that it resonates at f>20Hz we simply plug in f=20 Hz and find our n:
We can see that any resonant frequency is simply a multiple of a base frequency.
Let us find which harmonic resonates with the frequency 20 Hz:
Since n has to be an integer, final answer would be 323.
Where n is the resonance node, v is the speed of sound in air and L is the length of a cylinder.
The fundamental frequency is simply the lowest resonant frequency.
We find it by plugging in n=1:
To find what harmonic has to be excited so that it resonates at f>20Hz we simply plug in f=20 Hz and find our n:
We can see that any resonant frequency is simply a multiple of a base frequency.
Let us find which harmonic resonates with the frequency 20 Hz:
Since n has to be an integer, final answer would be 323.