Ask & Answer Question

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. New amplitude of oscillation will be:

Option 1 -

$A \sqrt[]{\frac{M}{M - m}}$

Option 2 -

$A \sqrt[]{\frac{M}{M + m}}$

Option 3 -

$A \sqrt[]{\frac{M - m}{M}}$

Option 4 -

$A \sqrt[]{\frac{M + m}{M}}$

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Velocity of block in equilibrium, in first case,

$v = A ω = A . \sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v' = A' ω' = A' \sqrt[]{\frac{k}{M + m}}$

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow M A \sqrt[]{\frac{k}{M}} = (M + m) A' \sqrt[]{\frac{k}{M + m}} \Rightarrow A' = A \sqrt[]{\frac{M}{M + m}}$

Velocity of block in equilibrium, in first case, <math> <mrow> <mi>v</mi> <mo>=</mo> <mi>A</mi> <mi>ω</mi> <mo>=</mo> <mi>A</mi> <mo>.</mo> <mroot> <mrow> <mfrac> <mrow> <mi>k</mi> </mrow> <mrow> <mi>M</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816697815"> </o:OLEObject></xml><! [endif]-> Velocity of block in equilibrium, is second case, <math> <mrow> <mi>v</mi> <mo>'</mo> <mo>=</mo> <mi>A</mi> <mo>'</mo> <mi>ω</mi> <mo>'</mo> <mo>=</mo> <mi>A</mi> <mo>'</mo> <mroot> <mrow> <mfrac> <mrow> <mi>k</mi> </mrow> <mrow> <mi>M</mi> <mo>+</mo> <mi>m</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816697816"> </o:OLEObject></xml><! [endif]->   From conservation of momentum, Mv = (M + m) v’ <math> <mrow> <mo>⇒</mo> <mi>M</mi> <mi>A</mi> <mroot> <mrow> <mfrac> <mrow> <mi>k</mi> </mrow> <mrow> <mi>M</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mrow> <mo> (</mo> <mrow> <mi>M</mi> <mo>+</mo> <mi>m</mi> </mrow> <mo>)</mo> </mrow> <mi>A</mi> <mo>'</mo> <mroot> <mrow> <mfrac> <mrow> <mi>k</mi> </mrow> <mrow> <mi>M</mi> <mo>+</mo> <mi>m</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>⇒</mo> <mi>A</mi> <mo>'</mo> <mo>=</mo> <mi>A</mi> <mroot> <mrow> <mfrac> <mrow> <mi>M</mi> </mrow> <mrow> <mi>M</mi> <mo>+</mo> <mi>m</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816697817"> </o:OLEObject></xml><! [endif]->

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question