How much energy is there in a string of length 90 meters whose mass density is 5.5 grams/meter, when the string is carrying a traveling wave with amplitude 1.45 meters and frequency 330 Hz, if wave velocity is 171 m/s?
How much power does it take to maintain this wave (Technical Physics and Principles of Physics students need not answer this question)?
Each particle in the string undergoes SHM with amplitude 1.45 meters and frequency 330 Hz. From the mathematic of SHM, or simply from the circular model, the maximum velocity of each particle is therefore
You will recall that the total energy of any mass in SHM is constant throughout the motion, with all the energy potential at the extreme points of the motion and all the energy kinetic at the equilibrium point, where the velocity is maximum. Since every particle is in SHM with the same frequency and amplitude, though different particles are at different points and have differing mixes of KE and PE, the total energy of every particle is equal to the KE it has at its maximum velocity. We can therefore state that the total energy of the wave motion in the string is the energy its total mass would have at the maximum velocity:
where vMax is the maximum SHM velocity of a particle.
The total mass of the string is simply 5.5 grams/meter * 90 meters * (1 kg / (1000 grams) ) = .495 kg. We have already found vMax. Thus the total energy of the string is
The power required to maintain the wave is the energy supplied per unit of time. Each time the wave travels the length of the string, the 2233444 Joules in the string will pass on beyond the end of the string (perhaps to another string, perhaps into some mechanism that makes use of the energy, perhaps dissipated into some medium--it doesn't matter where it goes, it just goes).
- power = 2233444 Joules / ( .5263158 sec) = 4243544 Joules / sec = 4243544 watts.
University Physics students note: To be completely rigorous we should consider a Riemann sum of the energies over small increments. A typical length increment `dx of the string will have its SHM defined by a sample point xi in the ith interval. Its total energy will be equal to that of the mass 5.5 g/m * `dx in the interval, and will hence be .5( 5.5 g/m * `dx ) * ( 3004 m/s)^2 = 4.963209E+07 kg / s^2 * `dx. When this quantity is summed over all intervals the 4.963209E+07 kg/s^2 factors out, the remaining increments `dx add up to the length of the string and we obtain 4.963209E+07 kg/s^2 * 90 meters = 2233444 Joules. This result does not depend on the increments `dx, but the approximation that all the mass in `dx moves with the same SHM is not completely valid. However, as the interval length `dx approaches zero, the approximation becomes more and more valid, and becomes exact in the limit.
A mass m moving in SHM with amplitude A and frequency f has angular frequency `omega = 2 `pi f, from which we see that its maximum velocity is `omega A = 2 `pi f A. Its kinetic energy at this maximum velocity is therefore
This is the same as the total energy of the mass at any point. Thus every mass increment of the string has total energy given by this expression, and when the resulting energies are totaled for the entire string we obtain
The total mass of the string is `mu * L, where L is string length and `mu is mass per unit length. Thus
If the wave moves with velocity v, the total energy of the wave will pass beyond the string in time
so we must supply energy at the rate