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blsea [12.9K]
2 years ago
14

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

held by the musculature in its socket.
If the mass of the eyeball is 7.5 g, a typical value, what is the effective spring constant of the musculature attached to the eyeball
Physics
1 answer:
STALIN [3.7K]2 years ago
4 0

The effective spring constant of the system is 39.6 N/m

Explanation:

The frequency of oscillation of a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k is the spring constant of the system

m is the mass

In this problem, we have:

f = 29 Hz is the frequency of vibration of the eyeball system

m = 7.5 g = 0.0075 kg is the mass

We can therefore re-arrange the equation to find the effective spring constant of the system. We find:

k=(2\pi f)^2 m = (2\pi (29))^2 (0.0075)=39.6 N/m

#LearnwithBrainly

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You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back
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Average speed: 86 km/h

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Driving way back:

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Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

f=340.9744 Hz

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