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Katyanochek1 [597]
3 years ago
10

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

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 that holds the eyeball in the socket?
Physics
1 answer:
Likurg_2 [28]3 years ago
4 0

Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

k=Spring constant

So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

33167.69 = 7.5 x 10⁻³  k

k=4422.35 x  10³ N/m

k = 4422.35  KN/m

Therefore spring constant will be 4422.35  KN/m

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Read 2 more answers
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is
olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

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

Total number of oscillation per unit time is called frequency of oscillation. Here, f=\dfrac{31}{15}=2.06\ Hz

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{15}{4\pi^2\times 2.06^2}

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

v_{max}=A\times \omega

v_{max}=A\times 2\pi f

v_{max}=0.075\times 2\pi \times 2.06

v_{max}=0.970\ m/s

v_{max}=97\ cm/s

Hence, this is the required solution.

5 0
3 years ago
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