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2 answers:
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ANSWER:100dB
f<em>rom the sound intensity level,the sound intensity is calculated as:</em>
<em /><em>₁=</em>
<em>=(10dB)㏒₁₀(l₁/l₀)</em>
<em>inserting numbers:</em>
<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>
<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>
<em>can be used to solve for l₁ and get</em>
<em> l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>
<em>since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>
<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>
<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>
<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>
<em> </em><em>₂=</em>
<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>
<em> </em><em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>
<em /><em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>
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Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2 √m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz