Question

1-state the relationship between a wave's velocity wavelength and frequency?

2- What is the principle of superposition?

3- The distance between two nodes or two antinodes in a standing wave is …………………. Wavelength?

4- - The distance between an antinode and a node in a standing wave is …………………. Wavelength?

5- A string, which has a mass of 0.50 kg and a length of 3.0 meters, is clamped on both ends. If the tension in the string is 25 Newtons, what are the lowest 3 natural frequencies of oscillation of the string? For each case, include a sketch of the standing wave pattern.

Answer 1

1.

The relationship between a wave's velocity wavelength and frequency is given by

velocity =frequency *wavelength

     v = f*lmada

2. This principle states that the resultant displacement of a particle of the medium acted upon by two or more waves simulataneously is the aglebraic sum of the displacements of the sameparticledue to individual waves, in the absence of the others.

According the superposition principle , the instantaneous resusltant displacement R of the particle due to two wavesactingtogether is expressed by

R =Y1+Y2

If the two displacements are in opposite directions ,the instanntaneous resultant displacement due to to two wavesacting together is expressed by

R =Y1-Y2

3.

The distance between two conjugative nodes or anti-nodes is λ/2.(Lamda/2)

4.

The distance between an antinode and a node in a standing wave is λ/4 (Lamda/4)

5.

Given that

A string has a mass (m) =0.50kg

The length of the string(L) =3.0m

Tension in the string is (T) =25N

The velocity of the wave is given by

                      v =Sqrt(T/(m/L)) =Sqrt(25/(0.50/3) =12.24m/s

The lowest frequency mode for a stretched string is called the fundamental, and its frequency is given by

                    f =Sqrt((T/(m/l))/2L =v/2L =12.24m/s/2*3 =2.04Hz

similarly we find the other two frequencies