Question

Both the strings, shown in figure (15-Q1), are made of same material and have same cross section. The pulleys are light. The wave speed of a transverse wave in the string AB is ν1 and in CD it is ν2. Then ν1/ν2 is
(a) 1
(b) 2
(c) √2
(d) 1/√2.
Figure

Answer 1

Both the strings, shown in figure (15-Q1), are made of same material and have same cross section. The pulleys are light. The wave speed of a transverse wave in the string AB is ν1 and in CD it is ν2. Then ν1/ν2 is 1/√2.

Explanation:

Suppose the system is maintained in equilibrium,

The tension over pulleys in the string is equivalent in every part say = T.  

So the string tension = 2T.  

Tension: Tension may be described as the force of pulling Forwarded axially through a string, A single-dimensional Ongoing cable, chain, or similar object, A single-dimensional continuous cable, chain, or the like; Tension could also be defined as the set of forces acts at each end of said elements. Compression can be the opposite of tension.

$\text { So } \frac{v^{1}}{v^{2}}=\sqrt{\left(\frac{F^{1} \mu^{2}}{F^{2} \mu^{1}}\right)}$

But $\mu_{1}=\mu_{2}$

$=\sqrt{\frac{\left(F_{1}\right)}{F_{2}}}=\sqrt{\left(\frac{T}{2 T}\right)}=\frac{1}{\sqrt{2}}$      

Answer 2

ν1/ν2 is (c) √2

The diagram given in the question is attached as image below.

Explanation:

Let the tension of the string CD = T

Let the tension of the string AB = T + T = 2T

Since, string AB is above CD, tension of both the strings are added.

The wave speed of a transverse wave in the string AB = ν1

The wave speed of a transverse wave in the string CD = ν2

The velocity of the wave in string is given by the formula:

$\nu=\sqrt{\frac{T}{\mu}}$

Where,

T = Tension

μ = Mass per unit length

The velocity of the wave in string AB is:

$\nu_{1}=\sqrt{\frac{2 T}{\mu}}$

The velocity of the wave in string CD is:

$\nu_{2}=\sqrt{\frac{T}{\mu}}$

Now, on substituting velocity ν₂ in ν₁, we get,

$\nu_1 = \sqrt{2} \ \nu_2$

Thus, the ration of velocity is given as:

$\frac{\nu_{1}}{\nu_{2}}=\frac{\sqrt{2}}{1}=\sqrt{2}$