How is Frequency related to Energy in a Wine Glass?

AI Thread Summary
Frequency and energy are related through quantum mechanics, but this relationship is not straightforward in classical physics. As energy increases, frequency typically increases, but exceptions exist, such as low-energy, high-frequency waves. In the context of a wine glass, filling it with liquid decreases the frequency of the sound produced because the water requires more energy to move, thus lowering the wave's energy. The frequency of a wine glass can also be affected by its effective vibrating area, which is reduced when water is added, leading to higher frequencies. Overall, the relationship between frequency and energy can vary depending on the physical context and medium involved.
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is there an equation that relates frequency to energy on a larger scale, relating to standing waves...

what proves that as energy increases, so does frequency of a wave?
 
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Quantum mechanics relates frequency to energy, but this strict relation is not present in classical physics. You can have low-energetic, high-frequency waves and high-energetic, low-frequency waves.
 
the reason that the frequency decreases in a wine glass as it is filled up with liquid is because it is using more energy to drag the water molecules with it, so the energy of the wave is decreasing, but so does frequency so they are directly related in this case?
 
when you talk about frequency of a wine glass, are you talking about a musical instrument like a glass organ? Acoustically you get a higher frquency when the glass has more water because the water reduces the glass area that can effectively vibrate to make the sound.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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