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Anettt [7]
4 years ago
5

Unpolarized light of intensity i0 is incident on three polarizing filters. the axis of the first is vertical, that of the second

is 43° from vertical, and that of the third is horizontal. what light intensity emerges from the third filter?
Physics
1 answer:
Eva8 [605]4 years ago
7 0
To determine the intensity, the key idea is to work through the system filter-by-filter and applying either the one-half rule for unpolarized light or the cosine-squared rule for an already polarized light.

When the light reach the first filter, it is unpolarized, so the intensity is :
            I1 = 1/2 I0                     (One-half rule)

Because the light reaching the second filter is polarized, given 43° angle difference, the intensity is : 
            I2 = I0 cos^2 (angle difference)
                = I0 cos^2 43°
                = 0.535 I0

Lastly, when the light reached the third filter, it is polarized and has angle diference equals to the angle between the second and last filter 47°. Using the same rule, we get :
           I3 = I2 cos^2 47°
               = (0.535 I0) cos^2 47°
               = (0.535 I0) (<span>0.465)
               = 0.249 I0

So, the intensity which emerges is 0.249 I0.</span>
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The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplit
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Solution :

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

$V_{max} = \omega A = 2 \pi f A$

Therefore, $f= \frac{V_{max}}{2 \pi A}$

$f= \frac{5}{2 \times 3.14 \times 2.5}$

f   = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.

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3 years ago
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot
FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0

with m = 1 slug and k = 9 lb/ft, the equation becomes

\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0
3 years ago
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