
Two wires of different densities but same area of cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.
Answer
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Hint: Velocity of transverse wave in wire is related to the tension and the mass per unit length of the wire. The wires are connected at one end, the assembly is in series and thus tension in both of them will be the same. Use this and the condition of velocity in question to find the ratio between densities.
Complete answer:
A transverse wave travels through the two wires at different speeds. Let the speed of waves be ${u_1},{u_2}$ in wires of density $\rho 1,\rho 2$ respectively. Both the wires have the same cross-sectional area and let that be $a$.
Both the wires are soldered at one end and tensed on the other. So the tension in both the wires is the same and let that be $T$.
If the length of the wire is ${l_1}$, then mass of the wire $ = {\rho _1}{a_1}{l_1}$ which is just a product of density and volume of the wire.
The speed of the wave through the wire $\sqrt {\dfrac{T}{{{\mu _m}}}} $, where ${\mu _m}$ is the mass of the wire per unit length. For first wire, mass per unit length $ = \dfrac{{{\rho _1}a{l_1}}}{{{l_1}}} = {\rho _1}a$ and therefore speed of wave through first wire $ = {u_1} = \sqrt {\dfrac{T}{{{\rho _1}a}}} $ (1)
Similarly for second wire, $ = {u_2} = \sqrt {\dfrac{T}{{{\rho _2}a}}} $ (2)
But they mentioned the relation between the values of speeds in each wire. The velocity of the transverse wave in the first wire is double that observed in the second wire. Using this relation with (1) and (2), we get,
$\Rightarrow {u_1} = 2{u_2}$
$\Rightarrow \sqrt {\dfrac{T}{{{\rho _1}a}}} = 2\sqrt {\dfrac{T}{{{\rho _2}a}}}$
$\Rightarrow {\rho _2} = 4{\rho _1}$
$\Rightarrow \dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{1}{4}$
Thus, we get a relation between the densities of the wires. The second wire is 4 times as dense as the first wire.
Note: When wires are soldered at one end, then this arrangement is analogous to series resistance connection in electrical systems where the current through all the resistance remains the same.
Complete answer:
A transverse wave travels through the two wires at different speeds. Let the speed of waves be ${u_1},{u_2}$ in wires of density $\rho 1,\rho 2$ respectively. Both the wires have the same cross-sectional area and let that be $a$.
Both the wires are soldered at one end and tensed on the other. So the tension in both the wires is the same and let that be $T$.
If the length of the wire is ${l_1}$, then mass of the wire $ = {\rho _1}{a_1}{l_1}$ which is just a product of density and volume of the wire.
The speed of the wave through the wire $\sqrt {\dfrac{T}{{{\mu _m}}}} $, where ${\mu _m}$ is the mass of the wire per unit length. For first wire, mass per unit length $ = \dfrac{{{\rho _1}a{l_1}}}{{{l_1}}} = {\rho _1}a$ and therefore speed of wave through first wire $ = {u_1} = \sqrt {\dfrac{T}{{{\rho _1}a}}} $ (1)
Similarly for second wire, $ = {u_2} = \sqrt {\dfrac{T}{{{\rho _2}a}}} $ (2)
But they mentioned the relation between the values of speeds in each wire. The velocity of the transverse wave in the first wire is double that observed in the second wire. Using this relation with (1) and (2), we get,
$\Rightarrow {u_1} = 2{u_2}$
$\Rightarrow \sqrt {\dfrac{T}{{{\rho _1}a}}} = 2\sqrt {\dfrac{T}{{{\rho _2}a}}}$
$\Rightarrow {\rho _2} = 4{\rho _1}$
$\Rightarrow \dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{1}{4}$
Thus, we get a relation between the densities of the wires. The second wire is 4 times as dense as the first wire.
Note: When wires are soldered at one end, then this arrangement is analogous to series resistance connection in electrical systems where the current through all the resistance remains the same.
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