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

(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`

Step-by-step 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{(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* 9.8`

`T=16.66\ N`

Now using eq. (1)

`v=\sqrt{(16.66)/(0.059) }`

`v=16.804\ m.s^(-1)`

(b)

Wavelength is given by:

`\lambda=(v)/(f)`

`\lambda=(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`