Question

A rope, under a tension of 272 N and fixed at both ends, oscillates in a second-harmonic
standing wave pattern.
The displacement of the rope is given by y = 0.40sin(πx/2.0) sin(13πt), where x = 0 at
one end of the rope, x and y are in meters, and t is in seconds.
The length of the rope is
A.1 m B. 2 m C. 4 m D. 8 m E. 16 m

The mass of the rope is in the previous question is
A. 1.61 kg B. 1.92 kg C. 2.33 kg D. 5.14 kg E. 8.97 kg

Answer 1

i get 4m for the length of the rope

PI/2=(2*PI)/landa
=4m

but thats wrong!

Answer 2

oh just noticed its a standing wave...

Answer 3

does that mean i multiply k by 2?

Answer 4

so then landa equals 2
2(PI/2)=(2*PI)/landa
PI*landa=2*PI
landa=2

Answer 5

lost for the next bit

Answer 6

ok i just realised i have 2 practice papers and one says 4m and the other says 2m. now i am really lost lol

Answer 7

if its second harmonic L=landa=(2*landa)/(landa)

so it still is landa
so i was right the first time...its 4m.

so one past paper is right.
am i right?

Answer 8

Length is 4 metres and mass is 1.61 kg.

Answer 9

ah k
so i just realised that \[\omega=\sqrt{k/m}\]
but i don't understand the next bit
\[m=T(k/\omega)^{2}L\]
what is the L?
that was in my lecture notes

Answer 10

it is supposed to be 4

Answer 11

can you show working and explanation of formulas used?

Answer 12

please

Answer 13

hmmm...

Answer 14

hmmm indeed

Answer 15

its a conspiracy

Answer 16

If you have second-harmonic, amplitude must be zero at x=0, x=L/2 and x=L.

Given sin(πx/2.0), this implies L=4.0m

so \(\lambda\)=4.0m and \(\omega\)=13\(\pi\)

Hence c=\(\lambda \nu=\Large\frac{\lambda\omega}{2\pi}\)=26 m/s

For a string: c=\(\Large \sqrt \frac T\mu\) so \(\mu=\frac {T}{c^2}\)=0.40 kg/m

M = Lµ = 1.6 kg