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Simple Harmonic Motion question

2015 · Shift 0 · Q60
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Simple Harmonic Motion question

2015 · Shift 0 · Q60

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
A pendulum made of a uniform wire of cross sectional area AAA has time period T.T.T. When an additional mass MMM is added to its bob, the time period changes to TM.{T_{M.}}TM.​ If the Young's modulus of the material of the wire is YYY then 1Y{1 \over Y}Y1​ is equal to : (g=gravitationalaccelerationg=gravitationalaccelerationg=gravitationalacceleration)
  1. A
    [1−(TMT)2]AMg\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}[1−(TTM​​)2]MgA​
  2. B
    [1−(TTM)2]AMg\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}[1−(TM​T​)2]MgA​
  3. C
    [(TMT)2−1]AMg\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}[(TTM​​)2−1]MgA​
  4. D
    [(TMT)2−1]MgA\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}[(TTM​​)2−1]AMg​
View written solutionFree

Correct answer: C

  1. Use the time period of a simple pendulum

For a pendulum of effective length LLL, time period is

T=2πLg.T = 2\pi \sqrt{\frac{L}{g}}.T=2πgL​​.

If the length changes to L′L'L′, then

TM=2πL′g.T_M = 2\pi \sqrt{\frac{L'}{g}}.TM​=2πgL′​​.

So,

(TMT)2=L′L.\left(\frac{T_M}{T}\right)^2 = \frac{L'}{L}.(TTM​​)2=LL′​.
  1. Relate the change in length to Young's modulus

Let the original length of the wire be LLL.

When an additional mass MMM is added to the bob, the extra force on the wire is

F=Mg.F = Mg.F=Mg.

Extension of a wire under force FFF is

ΔL=FLAY=MgLAY.\Delta L = \frac{F L}{A Y} = \frac{MgL}{AY}.ΔL=AYFL​=AYMgL​.

Hence the new length is

L′=L+ΔL=L(1+MgAY).L' = L + \Delta L = L\left(1 + \frac{Mg}{AY}\right).L′=L+ΔL=L(1+AYMg​).
  1. Substitute into the time period ratio

Since

(TMT)2=L′L,\left(\frac{T_M}{T}\right)^2 = \frac{L'}{L},(TTM​​)2=LL′​,

we get

(TMT)2=1+MgAY.\left(\frac{T_M}{T}\right)^2 = 1 + \frac{Mg}{AY}.(TTM​​)2=1+AYMg​.

Therefore,

(TMT)2−1=MgAY.\left(\frac{T_M}{T}\right)^2 - 1 = \frac{Mg}{AY}.(TTM​​)2−1=AYMg​.

Now solve for 1Y\frac{1}{Y}Y1​:

1Y=[(TMT)2−1]AMg.\frac{1}{Y} = \left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{A}{Mg}.Y1​=[(TTM​​)2−1]MgA​.
  1. Match with options

This matches:

[(TMT)2−1]AMg\left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{A}{Mg}[(TTM​​)2−1]MgA​

which is Option C.

  1. Check all options briefly
  • A: Gives negative value if TM>TT_M > TTM​>T, so incorrect.
  • B: Uses inverse ratio, incorrect.
  • C: Correct expression.
  • D: Dimensionally wrong for 1/Y1/Y1/Y.

So the correct answer is C.

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