Question

A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0⋅01 kg m−1 kept under a tension of 49 N. The fork produces transverse waves of amplitude 0⋅50 mm on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string?

Answer 1

Vmax  = 1.3816 m/s

amax =  3.8 km/s²

Average rate (p) = 0.67 W

Explanation:

Frequency of tuning fork f = 440 Hz

Linear mass density m = 0.01 kg/m

Applied tension T = 49 N

Amplitude of traverse wave = 0.5mm

Speed of the transverse wave v = root of (T/m)

v = root of (49/0.01) = 7/0.1 = 70 m/s

v = frequency / wavelength

So wavelength = frequency / speed = 70/ 440 = 16 cm

(a)  y = Asin(wt -kx)

v = dy/dt = Aw cos(wt -kx)

Vmax = dy/dt = Aw

= 0.5 * 10⁻³ * 2π * 440

= 1.3816 m/s

(b) a = d²y/ dt² = -Aw² sin (wt - kx)

amax = -Aw²

= 0.5 * 10⁻³ * 4π² * 440²

= 3.8 km/s²

(c) Average rate (p) = 2π²vA²f2

= 2 * 10 * 0.01 * 70 * (0.5 * 10⁻³)² * 440²

= 0.67 W

Answer 2

(a) The wave speed and the wavelength of the waves is $16 \mathrm{cm}$

(b) The maximum speed and acceleration of a particle of the string is $3.8 \mathrm{km} / \mathrm{s} 2$

(c) Average rate of the tuning fork transmitting energy to the string is$p=0.67 W$

Explanation:

It is given that ,

Turning fork has the frequency, $f=440 \mathrm{Hz}$

Linear mass density, $\mathrm{m}=0.01 \mathrm{kgm}-1$

The tension applied, $T=49 N$

Fork produce the amplitude of the transverse wave $=0.50$

The wave’s wavelength is denoted as $\lambda$

(a) The transverse wave has the speed that is shown as $\mathrm{v}=\mathrm{Tm}, \text { which is } 70 \mathrm{m} / \mathrm{s}$.  

Also, $v=70440=16 \mathrm{cm}$

(b) Maximum acceleration (amax) and maximum speed (vmax):

$\text { We have: } y=A \sin \omega t-k x$

$\therefore \mathrm{v}=\mathrm{d} y \mathrm{dt}=\mathrm{A} \omega$ $\cos \omega t-k x N o w$,$\text { vmax }=\mathrm{dydt}=\mathrm{A} \omega$$=0.50 \times 10-3 \times 2 n \times 4.40=$$=1.3816 \mathrm{m} / \mathrm{s} . \text { And }$$a=d 2 y d t 2 \Rightarrow a=-A \omega 2$

$\text { sin wt-kxamax }=-A \omega 2=0.50 \times 10-3 \times 4 n 2$ $4402=3.8 \mathrm{km} / \mathrm{s} 2$

(c) Average rate (p) is shown as

$p=0.67 \mathrm{W}$