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Akimi4 [234]
3 years ago
12

If all the strings on the guitar in the picture were plucked and allowed to vibrate freely, the string at the top would produce

the highest pitched sound. What could be concluded about the rate of vibration of the strings? The rates of vibration would vary over time for each string. There is no way to tell which string would be vibrating faster. The bottom string would be vibrating the fastest. The top string would be vibrating the fastest.
Physics
2 answers:
ivanzaharov [21]3 years ago
7 0

The top string would be vibrating the fastest

laiz [17]3 years ago
6 0

Answer: The top string would be vibrating the fastest.

Explanation: Here we do not have the picture, but we know that the string in the top would produce the highest pitched sound.

Now, the pitch of the sound is directly related to the frequency at which the string vibrates.

If the frequency is high, then the pitch is also high if the frequency is low, then the pitch is low.

now, let's analyze one of the sentences that are interesting.

"The rates of vibration would vary over time for each string."

As time passes, the intensity of the sound will decrease, not the frequencies.

Now, knowing that the top string is the highest-pitched one and that the high pitch is related with high frequencies, we can know that the top string is the one vibrating the fastest.

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2 years ago
A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t
Oduvanchick [21]

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just before landing the ground

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Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

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v = u

Thus, the velocity is same as it just reaches the ground.

6 0
3 years ago
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