Question

One end of a horizontal rope is attached to a prong of anelectrically driven tuning fork that vibrates at 120 Hz. The otherend passes over a pulley and supports a 1.70 kg mass. The linear mass density of the rope is0.0590 kg/m.(a) What is the speed of a transverse wave onthe rope?v1.70 kg =1 m/s(b) What is the wavelength?λ1.70 kg =2 m(c) How would your answers to parts (a) and (b) be changed if themass were increased to 2.80 kg?v2.80 kg = 3v1.70 kgλ2.80kg = 4λ1.70 kg

Answer 1

Answer:

(a) $v=16.804\ m.s^{-1}$

(b) $\lambda=1.4875\ cm$

(c)

• $F_T=27.44\ N$

• $v=21.57\ m.s^{-1}$

and

• $\lambda=17.97\ cm$

Explanation:

Given:

• frequency of vibration, $f=120\ Hz$

• mass of object attached to the rope, $m=1.7\ kg$

• linear mass density of rope, $\mu=0.059\ kg.m^{-1}$

(a)

We have the expression for velocity as:

$v=\sqrt{\frac{F_T}{\mu} }$ .................(1)

where:

$F_T=$ tension force in the rope

Now for tension force we balance the forces acting on the rope:

$T=m.g$

$T=1.7\times 9.8$

$T=16.66\ N$

Now using eq. (1)

$v=\sqrt{\frac{16.66}{0.059} }$

$v=16.804\ m.s^{-1}$

(b)

Wavelength is given by:

$\lambda=\frac{v}{f}$

$\lambda=\frac{16.66}{120}$

$\lambda=1.4875\ cm$

(c)

• On increasing the mass to 2.8 kg

We get :

$F_T=27.44\ N$

$v=21.57\ m.s^{-1}$

and

$\lambda=17.97\ cm$