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Waves question

2012 · Q169
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Waves question

2012 · Q169

NEETPhysicsWavesMCQ+4 / −1
When a string is divided into three segments of length lll1, lll2 and lll3 the fundamental frequencies of these three segments are υ1,υ2{\upsilon _1},{\upsilon _2}υ1​,υ2​ and υ3{\upsilon _3}υ3​ respectively. The original fundamental frequency (vvv) of the string is
  1. A
    v=v1+v2+v3\sqrt v = \sqrt {{v_1}} + \sqrt {{v_2}} + \sqrt {{v_3}}v​=v1​​+v2​​+v3​​
  2. B
    v=v1+v2+v3v = {v_1} + {v_2} + {v_3}v=v1​+v2​+v3​
  3. C
    1v=1v1+1v2+1v3{1 \over v} = {1 \over {{v_1}}} + {1 \over {{v_2}}} + {1 \over {{v_3}}}v1​=v1​1​+v2​1​+v3​1​
  4. D
    1v=1v1+1v2+1v3{1 \over {\sqrt v }} = {1 \over {\sqrt {{v_1}} }} + {1 \over {\sqrt {{v_2}} }} + {1 \over {\sqrt {{v_3}} }}v​1​=v1​​1​+v2​​1​+v3​​1​
View written solutionFree

Correct answer: C

Let lll be the length of the string. Fundamental frequency is given by

υ=12lTμ\upsilon = {1 \over {2l}}\sqrt {{T \over \mu }} υ=2l1​μT​​

⇒v∝1l \Rightarrow v \propto {1 \over l}⇒v∝l1​  (∵\because∵ T and μ\mu μ are constants)

Here, l1=kυ1,l2=kυ2,l3=kυ3{l_1} = {k \over {{\upsilon _1}}},{l_2} = {k \over {{\upsilon _2}}},{l_3} = {k \over {{\upsilon _3}}}l1​=υ1​k​,l2​=υ2​k​,l3​=υ3​k​ and l=kυ l = {k \over \upsilon }l=υk​

But l=l1+l2+l3l = l_1 + l_2 + l_3l=l1​+l2​+l3​

1υ=1υ1+1υ2+1υ3{1 \over \upsilon } = {1 \over {{\upsilon _1}}} + {1 \over {{\upsilon _2}}} + {1 \over {{\upsilon _3}}}υ1​=υ1​1​+υ2​1​+υ3​1​

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