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Pendulum, waves and diffraction grating questions

Hi was doin a multiple choice qstion paper and was a bit confused with these three qs. Help much Appreciated with explanation..

1> When the length of a simple pendulum is decresed by 600mm, the period of oscillation is halved. What is the original length of the pendulum?
A 800mm
B 1000mm
C 1200mm
D 1400mm

2> A wave of frequency 5HZ TRAVELS AT 8Kms through a medium. What is the phase difference, in radians, between two points 2km apart?
A 0
B PI/2
C PI
D 3PI/2

3> A narrow beam of monochromatic light falls on a diffraction grating at normal incidence. The second order diffracted beam makes an angle 45deg with the grating. What is the highest order visible with this grating at this wavelength?

A 2
B 3
C 4
D 5
t=2pi x sqrt (l/g)
t/2=2pi x sqrt ((l-0.6)/g)
Then try it out with the various values for l, you will find that it is in fact A.

C=f lamda
lamda = c/f
lamda = 8000/5 = 1600m = 1 wave = 2pi

2000-1600 = 400.
400/1600 = 1/4 of a cylce = pi/2

2pi + pi/2 = 3pi/2

I could proabably answer the last one if I was sober :smile: Btw, take my answers with a pinch of salt, seeing as i;m drunk :P
Reply 2
3.) 2nd order occurs at an angle such that

sin 45 = 2 lambda / d
so lambda /d = 0.35

3rd order would occur at

sin theta = 3 lambda / d = 3 x 0.35 . Not possible since this is >1 so 2nd order is the highest that can be observed.
Reply 3
2) calculate the wavelength lambda = c / f

The caclualte what fraction of a wavlength 2km is.
Original post by winnie the poo
t=2pi x sqrt (l/g)
t/2=2pi x sqrt ((l-0.6)/g)
Then try it out with the various values for l, you will find that it is in fact A.

C=f lamda
lamda = c/f
lamda = 8000/5 = 1600m = 1 wave = 2pi

2000-1600 = 400.
400/1600 = 1/4 of a cylce = pi/2

2pi + pi/2 = 3pi/2

I could proabably answer the last one if I was sober :smile: Btw, take my answers with a pinch of salt, seeing as i;m drunk :P


This is incorrect