A. Draw the wave that results when the two waves shown interact through destructive interference. (Image attached)
2 answers:
a) Since these waves are opposite, they will cancel one another (see fig.)
b) If we consider sound speakers, the waves may interfere and destruct each other (like on the attached figure), somewhere they will increase each other (constructive interference).
c) Echo. Go to mountains and scream (beware of avalanche), you'll wonder how many sound waves will come to you from different surfaces against you. This is used in echolocation.
PART A)
Since two waves are in opposite phase
So due to superposition of these type of opposite phase waves the resultant wave is of zero amplitude
This is known as destructive interference
PART B)
Sound from stage seem to be louder at certain seats because of constructive interference of sounds which are coming directly and coming after reflection from walls and other points
This superposition is the result where we get larger amplitude sue to superposition of same phase
PART C)
When a travelling wave bounce back in opposite direction by the change in medium boundary then this phenomenon is known as reflection of wave
If the reflected sound is in opposite phase then the superposition of such sound will give destructive interference
If reflected sound is in same phase then the resultant sounds is always constructive interference
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Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere