Answer:
0.21486 mm
Explanation:
The formula for the maximum intensity is given by;
I = I_o•cos²(Φ/2)
Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)
Where;
y is the distance from the central maximum
d is the distance between the slits
λ is the wavelength
L is the distance to the screen
Thus;
I = I_o•πdy/(λL)
We are given;
d = 0.05 mm = 0.5 × 10^(-3) m
λ = 540 nm = 540 × 10^(-9) m
L = 1.25 m
I/I_o = 50% = 0.5
From earlier, we saw that;
I = I_o•πdy/(λL)
We have I/I_o = 0.5
Thus;
I/I_o = πdy/(λL)
Plugging in the relevant values;
0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)
Making y the subject, we have;
y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))
y = 0.00021486 m
Converting to mm, we have;
y = 0.21486 mm
First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
Answer:
0 m/s
Explanation:
velocity= change in displacement/ time
at rest, the ball does not travel any distance
0/ t
=0
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
