# Homework Help: Problem regarding concave mirror

1. Jul 4, 2014

### sankalpmittal

1. The problem statement, all variables and given/known data

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.

2. Relevant 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

3. 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 ?

2. Jul 4, 2014

### Redbelly98

Staff Emeritus
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:
So what would dv/du be?

Last edited: Jul 4, 2014
3. Jul 4, 2014

### Staff: Mentor

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.

4. Jul 5, 2014

5. Jul 7, 2014

### Vibhor

Hello sankalpmittal

Is 8mm the correct answer ?

6. Jul 7, 2014

### Redbelly98

Staff Emeritus

7. Jul 9, 2014

### sankalpmittal

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

8. Jul 9, 2014

### Vibhor

Thanks

Last edited: Jul 9, 2014