PlanetMaster

wont

wont we subtract 0.94 from 1D as it says DECREASED BY 94%
It says decreases to 94%!
If it were decreases by 94%, then we would subtract 0.94 from 1.

redox233

CAN SOMEONE EXPLAIN THIS QUESTION?

redox233

AND THIS QUESTION AS WELL THANKYOU. HOW WILL WE FIND THE AMPLITUDE IN THIS QUESTION?

Last edited by a moderator:

PlanetMaster

View attachment 70518
CAN SOMEONE EXPLAIN THIS QUESTION?
You need to take intersections of both waves where they are going in the same direction.
From the graph, these can be either 0 to 0.5 or it can be 0.5 to 0.8.

We can then use ratios to calculate the phase difference
[imath]\frac{\Delta t}{T} = \frac{\Delta \phi}{360^{\circ}}[/imath]

Since T is 0.8 i.e one full cycle, using [imath]\Delta t = 0.3[/imath] gives us [imath]\Delta \phi = 135^{\circ}[/imath] and using [imath]\Delta t = 0.5[/imath] gives us [imath]\Delta \phi = 225^{\circ}[/imath]

So the answer is C i.e 135 which is also 360-225 i.e. 135

PlanetMaster

View attachment 70537

AND THIS QUESTION AS WELL THANKYOU. HOW WILL WE FIND THE AMPLITUDE IN THIS QUESTION?
During one oscillation, point A moves a distance of 80mm as given in the question.
And during this one oscillation, point A would return to its original position. Any point would move a distance = 4 time the amplitude during one oscillation, .

Remember, as defined above, the displacement is up and down while the wave travels to the right (in this case). Amplitude is the maximum displacement form the equilibrium position.

Consider the point A for example, which is originally on the minima. During one oscillation, it will move
1. a distance a from its current position at A to the equilibrium position,
2. another distance a from the equilibrium position to reach the maxima (crest),
3. another distance a to move from the maxima (crest) back to reach the equilibrium position again,
4. and finally, another distance a from the equilibrium position to return to its original position at A.

Thus, amplitude = 80/4 = 20mm

redox233

Sorry to disturb you again and again but can you pls tell that how the graphs are intersecting at 0.5 and 0.8
You need to take intersections of both waves where they are going in the same direction.
From the graph, these can be either 0 to 0.5 or it can be 0.5 to 0.8.

We can then use ratios to calculate the phase difference
[imath]\frac{\Delta t}{T} = \frac{\Delta \phi}{360^{\circ}}[/imath]

Since T is 0.8 i.e one full cycle, using [imath]\Delta t = 0.3[/imath] gives us [imath]\Delta \phi = 135^{\circ}[/imath] and using [imath]\Delta t = 0.5[/imath] gives us [imath]\Delta \phi = 225^{\circ}[/imath]

So the answer is C i.e 135 which is also 360-225 i.e. 13

PlanetMaster

Sorry to disturb you again and again but can you pls tell that how the graphs are intersecting at 0.5 and 0.8
No worries pal. For both waves, we can take the points they intersect 0 displacement in the same direction.

Like this:

Attachments

• Screenshot 2023-03-21 132418.png
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Lemniscates

PlanetMaster can u pls explain this uestion? thankyou!
Hey redox233, I am not sure if you still need help on this, but let me explain this to you:-

[imath]\rightarrow[/imath] This is a classic problem of standing waves.
[imath]\rightarrow[/imath] Given that the lowest sound produced is [imath]92[/imath] Hz, such as a node at the mouthpiece and an antinode at the other open end.

Hence, this is what we have,

We know that distance between a node and an antinode is [imath]\lambda/4[/imath].

[imath]\therefore \lambda/4[/imath] corresponds to [imath]92 \space \mathrm{Hz}[/imath].

The next arrangement of the standing wave to produce another frequency is as follows,

Here, we have [imath]\lambda/2 + \lambda/4 = 3\lambda/4[/imath]

[imath]\therefore 3\lambda/4[/imath] corresponds to [imath]3 \times \lambda/4 = 3 \times 92 =[/imath] [imath]276 \space \mathrm{Hz}[/imath]

Here is the next arrangement and you can follow this idea to get the overall correct answer to this question,

Here, we have [imath]\lambda + \lambda/4 = 5\lambda/4[/imath]

[imath]\therefore 5\lambda/4[/imath] corresponds to [imath]5 \times \lambda/4 = 5 \times 92 =[/imath] [imath]460 \space \mathrm{Hz}[/imath]

[imath]\rarr[/imath] If you simply look at all the options now in the question, we can clearly see that C seems to be the correct option.
[imath]\rarr[/imath] However, feel free to visualize the next arrangement and its corresponding frequency and match it with option C.

I hope this helps.
Good luck!

Lemniscates

No worries pal. For both waves, we can take the points they intersect 0 displacement in the same direction.

Like this:View attachment 70545

Hey PlanetMaster,

What software do you use to create such graphs that go with the background of the XPC theme? 🤔
Also, it would be great if you could provide a Math editor in the options so that we don't have to externally use LaTeX to write Mathematical equations here.

Thank you! ^_^

unknownerror21

a child of mass 35 kg movies down a sloping path on a skate board. the sloping path makes an angle 45 degree with the horizontal. the constant speed of the child along the path is 6.5 m/s. Calculate:
a) vertical dist through which child moves in 1.0s (done)
b) the rate at which potential energy is being lost (g=9.81 m/s^2) (confusion)

How is the answer to b) 80 J/s??

All I did was P = (35x9.81x0.51)/1 which gives answer 175 J/s (or 175 W)

a child of mass 35 kg movies down a sloping path on a skate board. the sloping path makes an angle 45 degree with the horizontal. the constant speed of the child along the path is 6.5 m/s. Calculate:
a) vertical dist through which child moves in 1.0s (done)
b) the rate at which potential energy is being lost (g=9.81 m/s^2) (confusion)

How is the answer to b) 80 J/s??

All I did was P = (35x9.81x0.51)/1 which gives answer 175 J/s (or 175 W)
A)
vertical distance moved: sin(45)=6.5/dist
dist = 0.7071*6.5
dist = 4.596

B)
rate at which potential energy is lost is just the energy change in one sec (i.e. vertical dist change in part A)
p.e = mgh = 35*9.81*4.596
p.e. = 1578 J/s

Hannah Moore

Hey everyone, if a person gets a wrong trend in the values in P3 physics Q.1. What is the procedure of marking for them.