A stretched string is forced to transmit transverse waves by means of an oscillator coupled to one end. The…
A stretched string is forced to transmit transverse waves by means of an oscillator coupled to one end. The string has a diameter of 4 mm. The amplitude of the oscillation is 10$^{-4}$ m and the frequency is 10 Hz. Tension in the string is 100 N and mass density of wire is 4.2 \times 10$^{3}$ kg/m$^{3}$. Find (i) the equation of the waves along the string, (ii) the energy per unit volume of the wave, (iii) the average energy flow per unit time across any section of the string, and (iv) power required to drive the oscillator.
Solution
Sol. (i) Speed of transverse waves in the string, v = $\sqrt{\dfrac{T}{\rho S}}$ (As, α = ρ S)
Substituting the given values in above equation, we get
v = $\sqrt{\dfrac{100}{(4.2 \times 10^3)(\pi/4)(4.0 \times 10^{-3})^2}}$ = 43.53 m/s
ω = 2πf = 20π rad/s = 62.83 rad/s (∴ f = 10Hz)
k = ω/v = 1.44 m$^{-1}$
∴ Equation of the waves along the string,
y (x, t) = A sin (kx − ωt) = 10$^{-4}$ sin (1.44 x − 62.83 t) m
(ii) Energy per unit volume of the string,
u = energy density = $\dfrac{1}{2} \rho \omega^2 A^2$
Substituting all the values in above equation, we get
u = $\left(\dfrac{1}{2}\right)(4.2 \times 10^3)(62.83)^2(10^{-4})^2$
= 8.29 × 10$^{-2}$ Jm$^{-3}$
(iii) Average energy flow per unit time,
P = Power = $\left(\dfrac{1}{2} \rho \omega^2 A^2\right)(Sv)$ = (u)(Sv)
Substituting the values, we get
P = (8.29 × 10$^{-2}$) (π/4) (4.0 × 10$^{-3}$)$^{2}$ (43.53)
= 4.53 × 10$^{-5}$ Js$^{-1}$
(iv) Power required to drive the oscillator is obviously 4.53 × 10$^{-5}$ W.
Answer: y(x,t) = 10$^{-4}$ sin(1.44 x − 62.83 t) m; energy density u = 8.29 × 10$^{-2}$ Jm$^{-3}$; average power flow P = 4.53 × 10$^{-5}$ Js$^{-1}$. Power required = 4.53 × 10$^{-5}$ W.