JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 13)
Two wires W1 and W2 have the same radius r and respective densities $$\rho $$1 and $$\rho $$2 such that ρ2 = 4$$\rho $$1 . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W1 to W2
is :
_8th_April_Morning_Slot_en_13_1.png)
1 : 1
1 : 2
1 : 3
4 : 1
Explanation
Fundamental frequency of a stretched string is ,
f = $${n \over {2L}}\sqrt {{T \over \mu }} $$
Here, n = number of antinodes
$$\mu $$ = mass per unit length.
As, O is the midpoint of two bridegs, hence length of two wires are equal.
$$\therefore\,\,\,$$ L1 = L2 = L
As frequency of both wires same
f1 = f2
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}} $$ = $${{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {4{\rho _1}}}} $$ = $${1 \over 2}$$
f = $${n \over {2L}}\sqrt {{T \over \mu }} $$
Here, n = number of antinodes
$$\mu $$ = mass per unit length.
As, O is the midpoint of two bridegs, hence length of two wires are equal.
$$\therefore\,\,\,$$ L1 = L2 = L
As frequency of both wires same
f1 = f2
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}} $$ = $${{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{n_1}} \over {{n_2}}}$$ = $$\sqrt {{{{\rho _1}} \over {4{\rho _1}}}} $$ = $${1 \over 2}$$
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