Question

A wire with mass 43.0 g is stretched so that its ends are tied down at points a distance 81.0 cm apart. The wire vibrates in its fundamental mode with frequency 59.0 Hz and with an amplitude at the antinodes of 0.350 cm.
a) What is the speed of propagation of transverse waves in the wire?
v = ______ m/s
b) Compute the tension in the wire.
F = _____ N
c) Find the magnitude of the maximum transverse velocity of particles in the wire.
= _____ m/s
d) Find the magnitude of the maximum acceleration of particles in the wire.
= m/s2

Answer 1

Answer:

Explanation:

mass, m = 43 g = 0.043 kg

L = 81 cm = 0.81 m

frequency, f = 59 Hz

Amplitude, A = 0.35 cm

(a)

Frequency

$f=\frac{v}{2L}$

v = 59 x 2 x 0.81 = 95.58 m/s

(b)

Let F be the tension in the string

$v=\sqrt{\frac{F}{\mu }}$

where, μ is mass per unit length

$\mu =\frac{0.043}{0.81}=0.053 kg/m$

$95.58=\sqrt{\frac{F}{0.053 }}$

$F = 95.58\times 95.58 \times 0.053$

F = 484.18 N

(c)

Maximum velocity

v = ω A = 2 π f A

v = 2 x 3.14 x 59 x 0.0035 = 1.3 m/s

(d)

Maximum acceleration

a = ω² A

a = (2 π f )² x 0.0035

a = ( 2 x 3.14 x 59)² x 0.0035

a = 480.5 m/s^2