370
8 Waves
8.29
5
2
λ = 15 cm
∴ λ = 6 cm
f =
v
λ
=
2400
6
= 400 Hz
ω = 2π f = 2512 rad/s
k =
2π
λ
= 1.047/cm
A = 6 cm
y = A sin(ωt − kx) = 6 sin(2512t − 1.047x) cm
8.30 P =
1
2
ω
2 A
2
μv = 2π
2 f
2 A
2
μv
= 2π
2
× (400)
2
(0.06)
2
(2.5 × 10
−4
)(24) = 68.2 W
8.31 (a) The amplitude of any point of the plucked string at time t may be
written as
y =
∞
n=1
a n cos ω n t sin
nπ x
L
+
∞
n=1
b n sin ω n t sin
nπ x
L
(1)
The kinetic energy of vibration of an element of length of string dx in the
nth mode is given by
dK n =
1
2
(μdx)( ˙
y)
2
=
1
2
μω
2
n (−a n sin ω n t + b n cos ω n t)
2 sin
2
(k n x)dx
(2)
where we have used the value of velocity ˙
y by differentiating (1) for the
nth mode with respect to t.
The potential energy of an element of string of length dx is
dU n =
1
2
ky
2 dx
=
1
2
μω
2
n (a n cos ω n t + b n sin ω n t)
2 sin
2 k n x dx
(3)
where we have used (1).
Adding (2) and (3), the total energy
dE n = dK n + dU n =
1
2
μω
2
n (a
2
n + b
2
n ) sin
2 k n x dx
(4)