Recent content by orangephysik

Oh okay. Thank you :smile:

Oh right. So ##s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)## ⇔##s_{1} + s_{2} =7 cm \cdot cos(27.5s^{-1}\cdot t-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x)## With ##cos(\alpha - \beta)=cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)##...

Oh right, that should be ##cos(0.075 m^{-1} \cdot x)## . I forgot the units. So ##s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)## and so the maximum would be at when ##cos(0)=1##. I get ##x_{max_{1}} = \frac{55s^{-1}\cdot t}{11.15 m}## and...

##\mathbf {Homework ~Statement:}## Consider the superposition of two one-dimensional harmonic waves $$s_1(x,t)=3.5 cm \cdot cos(27.5s^{-1} \cdot t - 5.65m^{-1} \cdot x)$$ $$s_2(x,t)=3.5 cm \cdot cos(27.5s^{-1} \cdot t - 5.5m^{-1} \cdot x)$$ ##\mathbf {a)}## Calculate the wavelength ##\lambda##...

I can't comprehend the logic behind this, could you explain how you did that? And is 167.0 s the correct answer? :) (Well technically 167.2 s if I plug in the exact values, and I get the same value too if I use the exact value of γ when solving 2=Φ_0 * exp(-γt) ).

I got t = 167.9 s. So I guess by omitting the cos term), we only consider the part when the pendulum swings back to its neutral position for the first time (I can see now why my assumption was false). But t =167.9 s doesn't seem right, since it only takes 1/f = 1/(0,27 Hz) = 3.70 s for one...

Hi, so of course Φ0 = 15° and after solving after solving Φ(t=5*T = 5/f) I found γ = 0.012 I need help with b). If I do 2° = 15° * exp(-0.012t)*cos(2πf*t), I'm not able to find t so I did something else by assuming that the amplitude decreases at a constant rate: After 5*T = 5*1/f = 18.52 s...

Oh right. The period T is the time it takes for the sound waves to travel a distance of λ - λ' = 0.3298 m. So since acceleration is a = Δv/Δt, I already have Δv since v_0 = 0 m/s and now I just need Δt. But the question implies that I can find the acceleration without knowing Δt. I have v(t) =...

Hi, I've attached the photo of the diagram, a photo of my drawings on the diagram. for a): Since the prism is an equalateral triangle, all angles inside the prism is 60°. This means the angle adjacent to α is 180° - 60° = 120°, which means the last angle is 180° - 24.5° - 120° = 35.5°. The...

I had f_0 = v_phasefront / λ and got λ = (343 m/s) / (520 Hz) = 0.6596 m, and this is the wavelength of the source when it was at rest. With the same formula, f_1 = v_phasefront / λ' , so λ' = (343 m/s) / (1040 Hz) = 0.3298 m, this is the wavelength when the source reaches me. Now using...

Hi. I need help with part a). I calculated the wavelength of the source by using the formula f_0 = v_phasefront / λ and got λ = (343 m/s) / (520 Hz) = 0.6596 m. And then I set up an equation for the velocity of the source v(t) = a*t (with v(t = 0 )= 0 m/s) and s(t) = 1/2 * at^2 + s_0. But I...