Thursday

July 24, 2014

July 24, 2014

Posted by **Lauren** on Sunday, August 5, 2012 at 9:31pm.

a.) What is the equation of the standing wave?

b.) At what distances from x=0 are the nodes and antinodes?

c.) What is the frequency of a point on the string at an antinode?

d.) If the string is 4m long, how many nodes are there?

- physics -
**Elena**, Monday, August 6, 2012 at 10:10amy(x,t) =|2•A•coskx|•cosωt,

y(x,t) =|2•0.005•cos(0.750•π•x)| •cos(942•t),

Antinods:

x=m• (λ/2), m=0,1,2,...

The 1st antinode at the origin (m=0),

the 2nd antinode at λ/2 (m=1).

Distance from origin = λ/2

Nodes:

x= (m+½)• (λ/2).

The 1st node at λ/4 (m=0),

the 2nd antinode at 3λ/4 (m=1)

Distance from origin = λ

The frequency is f=ω/2π=942/2 π =149.9 Hz.

λ=2π/k=2π/0.75 π=8/3 (meters).

If the antinode is at the origin, then we have

3 nodes on the distance 4 m at the points:

x=λ/4, 3λ/4, 5λ/4,

and 4 antinodes at

x= 0, λ/2, λ, 3λ/2.

- physics -
**Elena**, Monday, August 6, 2012 at 10:11amy(x,t) =|2•A•coskx|•cosωt,

y(x,t) =|2•0.005•cos(0.750•π•x)| •cos(942•t),

Antinods:

x=m• (λ/2), m=0,1,2,...

The 1st antinode at the origin (m=0),

the 2nd antinode at λ/2 (m=1).

Distance from origin = λ/2

Nodes:

x= (m+½)• (λ/2).

The 1st node at λ/4 (m=0),

the 2nd antinode at 3λ/4 (m=1)

Distance from origin = λ

The frequency is f=ω/2π=942/2 π =149.9 Hz.

λ=2π/k=2π/0.75 π=8/3 (meters).

If the antinode is at the origin, then we have

3 nodes on the distance 4 m at the points:

x=λ/4, 3λ/4, 5λ/4,

and 4 antinodes at

x= 0, λ/2, λ, 3λ/2.

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