More on fundamental frequency

[Soaring Crane]

A horizontal string of length 1.5 m vibrates with a wave velocity of 1320 m/s at its fundamental freq.

a. What is the fund. freq.?

Do I use the formula f_1 = v/4L?

b. What is the freq. of the 4th overtone and how many nodes and antinodes will it have?

To find 4th's freq. use f_n = 4*f_1?

Since this is 4th overtone, are there 6 nodes and 5 antinodes?

(assuming for 4th 5 positive crests on basic diagram for all situations)

c. In terms of interference to what do the nodes and antinodes correspond?

this is conceptual, right?

nodes - locations where string appears motionless due to destructive interference

antinodes - where string has greatest displacement due to constructive displacement.

d. how many g must string be to have a tension of 10^5 N?

Just use v = sqrt[F/(m/L)]

 

[Doc Al]

a. What is the fund. freq.?

Do I use the formula f_1 = v/4L?

No. The general formula is $f_m = m v/2L$.

b. What is the freq. of the 4th overtone and how many nodes and antinodes will it have?

To find 4th's freq. use f_n = 4*f_1?

The 4th overtone is the 5th harmonic (m = 5 in the above formula).

Since this is 4th overtone, are there 6 nodes and 5 antinodes?

Just by accident, this is correct.

c. In terms of interference to what do the nodes and antinodes correspond?

this is conceptual, right?

nodes - locations where string appears motionless due to destructive interference

antinodes - where string has greatest displacement due to constructive displacement.

Right (but that's constructive interference).

d. how many g must string be to have a tension of 10^5 N?

Just use v = sqrt[F/(m/L)]

Right.

 

[wais]

a. Using the formula f_1 = v/4L, the fundamental frequency of the string is f_1 = 1320/4(1.5) = 220 Hz.

b. To find the frequency of the 4th overtone, we can use the formula f_n = 4*f_1, which gives us f_4 = 4*220 = 880 Hz. The 4th overtone will have 6 nodes and 5 antinodes. This is because for the 4th overtone, there will be 5 positive crests on the string, which correspond to 5 antinodes, and 6 nodes in between each antinode.

c. The nodes and antinodes correspond to points of constructive and destructive interference respectively. At the nodes, the waves traveling in opposite directions on the string will cancel each other out, resulting in no net motion. At the antinodes, the waves will add up and produce the maximum displacement of the string.

d. To find the mass of the string needed to have a tension of 10^5 N, we can use the formula v = sqrt[F/(m/L)]. Solving for m, we get m = F*L/v^2. Plugging in the values, we get m = (10^5 N)(1.5 m)/(1320 m/s)^2 = 0.085 kg. Therefore, the string must have a mass of 0.085 kg to have a tension of 10^5 N.