Answer:
On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means
This means that the relation between the wavelength and the length of the string is
By definition, this standing wave is at the third harmonic, n = 3.
Furthermore, the standing wave equation is as follows:
The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.
For this equation to be equal to zero, sin(59.94t) = 0. So,
This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
<span>Is the following sentence true or false? Newton's first law does object's mass concentration and its axis of rotation increases, its rotational inertia The bicycle wheels at rest have no angular momentum, and the bicycle will fall over easily.</span><span>
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Answer:
The vibrational frequency of the rope is 5 Hz.
Explanation:
Given;
number of complete oscillation of the rope, n = 20
time taken to make the oscillations, t = 4.00 s
The vibrational frequency of the rope is calculated as follows;
Therefore, the vibrational frequency of the rope is 5 Hz.
