Question

Why 4??
If Amplitude = Distance/ Frequency..
And the frequency = 1/0.2 = 5 Hz..
Question in attached file thanks in advance.

Answer 1

Oh and the question is (ii) not (i)

Answer 2

What is the given answer ?

Answer 3

The given answer is =
80/4= 20 mm it's in the attached file :)

Answer 4

Ok, the time for one oscillation = 0.2 i.e Time period = 0.2, so frequency should be equal to 5 Hz.

Now as u said \(\sf \large Amplitude =\frac{Distance}{Frequency}\)
=> Amplitude = 80/5=16 mm

Answer 5

Aww yep That is what I did too but that is not the correct answer, the answer sheet states that \[80\div4=20\]

Answer 6

How frequency will be 4 ?
I think the answer given is wrong...are u sure that answer is true ?

Answer 7

Aww I know , that is what confused me too, and here is the answer sheet

Answer 8

ok, give a minute then

Answer 9

Awww of course and thanks a lot (: @Abhisar

Answer 10

@midhun.madhu1987

Answer 11

No idea... :(

Answer 12

it's ok..thanx for bothering :)

Answer 13

@Mashy

Answer 14

Aww thanks a lot everyone, once school starts I will ask my physics teacher and once I get the solution I will let you know through here!

Answer 15

@nincompoop

Answer 16

@aaronq

Answer 17

NOTE to all @Abhisar @MonsterIsEnergy.

You have confused amplitude with wavelength.

it is WAVELENGTH = velocity/f.

As I understand it amplitude is the maximum displacement from the mean position.

Since the point A moves 80mm (that is the peak to peak distance) I think the answer should be 80/2 = 40mm

Answer 18

OK - I made a mistake above.

The point A moves up AND down during 1 oscillation - so that is 2 * peak to peak , or 4 times amplitude.

So amplitude = 80/4 = 20

Answer 19

@MrNood oh that makes sense.
Thanks a lot.