Question

Hlo bhartvasiyon !

Happy republic day ☺

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A partical of mass 0.50 kg
executes a simple harmonic motion under a force F = -(50Nm-¹)x . if it crosses the centre of oscillation with a speed of 10ms-¹ find the amplitude of the motion

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Answer 1

Hello Bhartiya !
Happy Republic Day !

Your Solution ⏬

Energy at the mean setting is completely Kinetic and at the Full amplitude it's potential.

It may be solved as ;-

Work done in moving a small distance dx against the force is  = $<b> -F dx=50 x dx </b>$

Let A be the amplitude then unifying the function within 0 to A we get, <b>1/2(50 A²) is the PE now it should be equal to the KE as the total energy is conserved in SHM.

So, solving we get $<b>(0.5)(10)²=50 A²</b>$

#BeBrainly

Answer 2

$\bold \blue{hello!!}$

$\bold\green{solution}$
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:- The kinetic energy of the partical when it is at the centre of oscillation is E = 1/mv²

= 1/2 (0.50kg) ( 10ms-¹)²

= 25J

The potential energy is Zero here , at the maximum displacement X = A , speed is Zero and hence , kinetic energy is Zero the potential energy here is 1/2kA² . As there is no loss of energy .

1/KA² = 25J

the force on the partical is given by

F = - (50Nm-1 )x

this the spring constant is k = 50Nm-¹

Equation , (1( gives

1/2 ( 50N m-¹ ) A² = 25J

or, A = 1 Amplitude = 1m Answer ✔

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$\bold \orange{hope \: it \: helps}$

$\bold\orange{jai \: hindi}$
$\bold\pink{@bhartvasi \: rajuraj}$