<p>@recyclingbin @SheepLionWut </p>
<p>**Edit: I got a different answer from @SheepLionWut.</p>
<p>Originally, the wavelength is 2L, and frequency is “f”, therefore v = 2L*f.
Now, wavelength is 3L/2. wavespeed is still the same, at v, therefore:</p>
<p>2L*f = 3L/2 * f’
2f = 3/2 * f’
(2f / (3/2)) = f’
f’ = 4/3 f</p>
<p>@SheepLionWut @tamabombXD Isn’t the equation for frequency f = nv/(2L)? Therefore, if the node distance is 3/4 of the original distance, then the total L is also 3/4 of the original L, and so f = nv / 2 ( 3/4L) = 4/3 * nv/(2L)? Or am I missing something?</p>
<p>I just looked over waves this morning, so I’m probably not in a very good shape as well haha</p>
<p>Wait, but the original fundamental wavelength is 2x(L)=2L…so the new wavelength is still 3/4 the original…so the answer is 4f/3??? i guess my assumption is that velocity stays the same but don’t you have to assume that? don’t know much about waves tbh but i’m sticking with 4f/3</p>
<p>I’m pretty sure the answer should be 4/3 for the vibrating string question</p>
<p>The f=nv/(2L) is the equation, where n stands for whatever ‘harmonic number’ it is. However, when using this equation to determined frequency, the length of the string remains constant. Since the string is changing to 3L/4, must be change to f=nv/(2(3L/4)). I don’t know if i’m even right, but do you understand how changing the L makes it essentially a completely different problem</p>
<p>But velocity is constant. So if wavelength is decreased by a factor of 3/4, the frequency must increase to balance that out. I think velocity depends only on the properties of the string, like linear mass density and tension or something like that,so velocity is the same.</p>
<p>Using that equation, the original frequency is f=nv/2L=(1/2)nv/L and the new frequency is f=nv/2(3/4L)=(2/3)nv/L…however you must write this in terms of the original frequency. 2/3 is 4/3 of 1/2, so it’s 4/3.</p>
<p>Think it is 4/3 after seeing tama’s thing</p>
<p>what did everyone get for the negatively charged rod that was being grounded ?</p>
<p>@SheepLionWut What I’m trying to say is that even if the node distance is half of a wavelength, if you make all the node lengths 3/4 of the original node lengths, the total length L is 3/4 of the original L</p>
<p>Example: Take “-” to be one unit and | to be nodes (hence original node distance is 4)
Original: ----|----|----|----
New: —|---|—|---</p>
<p>Even though the node distance is 1/2 of a wavelength, the new total length L is still 3/4 of the original L. It’s like taking L = (4 + 4 + 4 + 4), and multiplying each of them by 3/4: L’ = (3/4 * 4 + 3/4 * 4 + 3/4 * 4 + 3/4 * 4); pulling out the 3/4 as a common factor we get L’ = 3/4 ( 4 + 4 + 4 + 4) = 3/4 (L). If we do it with full wavelengths, we get L = (8 + 8), but multiplying by 3/4 still gets us L’ = (3/4 * 8 + 3/4 * 8) = 3/4 ( 8 + 8 ) = 3/4 ( 4 + 4 + 4 + 4 ) = 3/4 (L).</p>
<p>More generally speaking, if L = (x + x + x + x + …), where x is the node distance, then we also have L = ( d + d + d + d + …), where d is the wave length and d = 2x. Either way, multiplying by 3/4 gets us 3/4L.</p>
<p>@sparkl3 What were the answer choices? I think I was a bit confused on that question.</p>
<p>yea it’s definitely 4/3 ■■■■</p>
<p>@sparkl3 that one was a bit confusing. They never explicitly said that the grounding cable had been removed so I just said the net charge on RS was 0. </p>
<p>sparkl3 the neutral object was grounded, answer was it became positively charged.
Reasoning: Brought a neg. rod towards it, force all electrons to opposite side of neutral object.
when neutral object was grounded, the electrons left the neutral object and went to ground, and after the object is removed from ground it would then have a positive charge due to the loss of its electrons</p>
<p>@thesoxpride10 wouldn’t there be a net positive charge on one end, as a result of polarization? I think that was one of the answer choices.</p>
<p>oh crap thesox you have a point, i just assumed it was removed otherwise it would be a ■■■■■■■■ question</p>
<p>Yeah, I put positively charged and I like @SheepLionWut’s reasoning. That one was tricky though. Not positive it was positive.</p>
<p>Does anyone have a list of the answer choices? I’m not sure which one I put at this point…</p>
<p>I only remember two: the net charge is 0 and the charged rod loses its charge</p>
<p>@SheepLionWut agree that the neutral object becomes positive.</p>
<p>ok, when a negatively charged rod is brought close to the neutral object (let’s assume that the neutral object is initially NOT grounded), then all the electrons in the neutral object shift away from the negatively charged rod (basically charging by induction).</p>
<p>however, when the neutral object becomes grounded, the electrons inside the neutral object leave through the Earth connection (due to presence of negatively charged rod, and like-like charge repulsion) leaving the <em>entire</em> object positive.</p>
<p>One of them was became negatively charge which makes 0 sense, the other was net charge of 0 which one could make the argument for if the object is ‘permanently grounded’ and the other two options i forget. It’s either positive or 0 but in hindsight that definitely should have specified whether or not it remains grounded</p>
<p>The other two choices had to do with positive charges moving. They can’t be true because only electrons move.</p>