A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and
1 answer:
To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
Therefore the number of wavelengths of the wave fit on the string is 6.
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Answer:
Stay the same
Explanation:
First of all, let's find how the capacitance of the capacitor changes.
Initially, it is given by
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:
The potential difference across the capacitor is given by
where
Q is the charge on the plates
C is the capacitance
We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:
Now we can analyze the electric field between the plates of the capacitor, which is given by
we said that:
- The voltage has doubled: V' = 2 V
- The distance between the plates has doubled: d' = 2 d
therefore, the new electric field will be
So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:
F = qE
Then the force on the particle will stay the same.