Problem regarding concave mirror

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Homework Help Overview

The discussion revolves around a problem involving a point object oscillating in front of a concave mirror. The object is positioned at 15 cm from the mirror, which has a radius of curvature of 20 cm. The task is to determine the amplitude of the image produced by the mirror as the object oscillates with a given amplitude.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the motion is simple harmonic motion (SHM) or another type of oscillatory motion. There are discussions about using the mirror formula and differentiating it to find relationships between object and image distances. Some participants question the appropriateness of approximations for small amplitudes and suggest focusing on maximum and minimum object distances.

Discussion Status

The discussion includes various attempts to analyze the problem, with some participants providing hints and alternative approaches. There is no explicit consensus on the solution, but guidance has been offered regarding the focus on object distance ranges rather than velocities.

Contextual Notes

Participants mention the need to consider the maximum and minimum distances of the oscillating object, indicating that assumptions about the motion's nature may affect the approach taken. There is also a reference to the potential pitfalls of using scattered study materials.

sankalpmittal
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Homework Statement



A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along principal axis with amplitude 2 mm. Find the amplitude of its image.

Homework Equations



Mirror formula :

1/v + 1/u = 1/f

Here u=-15 cm , f=-10

Differentiating above formula

dv/dt = -m2 du/dt

Where m= transverse magnification=-v/u

The Attempt at a Solution



Ok I don't know if it's SHM or just another sort of oscillatory motion. Had it been SHM I would have solved it by putting maximum velocity = amplitude times angular frequency.

SHM: Simple harmonic motion.

My attempt :

1/v + 1/u = 1/f
u=-15 cm , f=-10
v=-30

Thus m=-v/u = -2

Thus

dv/dt = -4 du/dt

Now ?

Please help !

Thanks in advance...

:smile:
 
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Tedius solution method (ignore unless totally stuck):
[STRIKE]What is the range of (i.e. maximum and minimum) object distances for the object, as it is oscillating? That should help get you started.[/STRIKE]

Quicker solution method:
sankalpmittal said:
dv/dt = -4 du/dt
So what would dv/du be?
 
Last edited:
Redbelly98 said:
Quicker solution method:
Maybe quick (I doubt that), but certainly wrong, this would be an approximation for small amplitudes. The amplitude is not large, but there is no reason to use an approximation here.

What is the range of (i.e. maximum and minimum) object distances for the object, as it is oscillating? That should help get you started.
That's the right approach.

You don't need velocities, or details of the motion. Just the maximal and minimal distance.
 
mfb said:
Maybe quick (I doubt that), but certainly wrong, this would be an approximation for small amplitudes. The amplitude is not large, but there is no reason to use an approximation here.

That's the right approach.

You don't need velocities, or details of the motion. Just the maximal and minimal distance.


Thanks a lot mfb and Radbelly. I got the answer.

However here Radbelly's hint had worked.
 
Hello sankalpmittal

Is 8mm the correct answer ?
 
Vibhor said:
Hello sankalpmittal

Is 8mm the correct answer ?
Please show your work.
 
Vibhor said:
Hello sankalpmittal

Is 8mm the correct answer ?


Yup. Well if you are jee aspirant I will advise you not to study from scattered sources.
 
Thanks
 
Last edited:

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