Respuesta :
Answer:
(a) [tex]v=16.804\ m.s^{-1}[/tex]
(b) [tex]\lambda=1.4875\ cm [/tex]
(c)
- [tex]F_T=27.44\ N[/tex]
- [tex]v=21.57\ m.s^{-1}[/tex]
and
- [tex]\lambda=17.97\ cm [/tex]
Explanation:
Given:
- frequency of vibration, [tex]f=120\ Hz[/tex]
- mass of object attached to the rope, [tex]m=1.7\ kg[/tex]
- linear mass density of rope, [tex]\mu=0.059\ kg.m^{-1}[/tex]
(a)
We have the expression for velocity as:
[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex] .................(1)
where:
[tex]F_T=[/tex] tension force in the rope
Now for tension force we balance the forces acting on the rope:
[tex]T=m.g[/tex]
[tex]T=1.7\times 9.8[/tex]
[tex]T=16.66\ N[/tex]
Now using eq. (1)
[tex]v=\sqrt{\frac{16.66}{0.059} }[/tex]
[tex]v=16.804\ m.s^{-1}[/tex]
(b)
Wavelength is given by:
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{16.66}{120}[/tex]
[tex]\lambda=1.4875\ cm [/tex]
(c)
- On increasing the mass to 2.8 kg
We get :
[tex]F_T=27.44\ N[/tex]
[tex]v=21.57\ m.s^{-1}[/tex]
and
[tex]\lambda=17.97\ cm [/tex]
