1.17 Standing Waves on a String of Length L When looking at standing waves we assumed that medium was of infinite extent. What happens if the medium has a fixed length? We shall consider a perfect string of length L, and mass per unit length µ,that is under a tension F. Each end of the string is fixed. We will excite a harmonic wave of amplitude A, at a frequency f, and wave speed v, on the string. This wave is reflected at each end of the string to form a standing wave.

Standing Waves on a String of Length L As each end of the string is fixed then we must have a node at each end. L x = 0 x = L Node How does the presence of the nodes affect the system? A node implies no displacement. Hence y(x=0,t) = y(x=L,t) = 0.

Lets us consider the equation for a standing wave When x = 0 then it is obvious that y(x=0,t) = 0. What about when x = L? For y(x=L,t) = 0 then sin(kL) = 0 This is satisfied when kL = nπ (n an integer 1,2….) As a result only certain wavelengths can be excited on the string. These wavelengths are given by

As the wavelength and frequency are linked then only certain frequencies can be supported on the string. These are These allowed frequencies are referred to as MODES OF VIBRATION or simply MODES. They correspond to frequencies at which energy is efficiently coupled to the string. We can excite modes in many different systems Guitar strings Laser cavities Atomic systems

Let us now look at how these modes behave. We know Now let us consider the case n = 1 (n = 0 does not make sense) This corresponds to the longest wavelength and lowest frequency that can be supported by the string and is known as the FUNDAMENTAL FREQUENCY or the FIRST HARMONIC of the string. For n = 2 we have This is the SECOND HARMONIC of the string.

Second Harmonic Third Harmonic Tenth Harmonic First Harmonic L

What is the frequency difference between successive harmonics or modes? Let the n th harmonic be frequency f n Let next harmonic be the (n+1) th harmonic be frequency f n+1 Let the difference in frequency be ∆f Each time the harmonic number changes by 1 the frequency changes by an amount equal to the fundamental frequency.

Example: Fundamental frequency of a guitar string. A guitar string has a mass per unit length of 0.4 g/m a length of 0.8 m and is placed under a tension of 50 N. What is the fundamental frequency of the string? The allowed frequencies of the string are given by With v the wave speed on the string, l the length of the string and n an integer. For the fundamental frequency n = 1.

Example: Fundamental frequency of a guitar string. The wave speed on the string is given by With F the tension and µ the mass per unit length. So the fundamental frequency of the string is given by Now F = 50 N µ = 4 x kg m -1 and l = 0.8 m. So f 1 = 221 Hz.

Example: Mode spacing in a laser cavity. A laser comprises two mirrors separated by 1 m. The light that oscillates between the two mirrors sets up a standing wave pattern. If the speed of light is 3 x 10 8 ms -1 what is the spacing between successive modes of the the cavity? We know that the mode spacing ∆f is given by Here l = 1 m and v = 3 x 10 8 ms -1 So ∆f = 150 MHz.