If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position?

Question

If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position? (A)  0.375 π 2 texterg  (B)  3.75 π 2 texterg  (C)  37.5 π 2 texterg  (D)  375 π 2 texterg

Tardigrade Admin · Accepted Answer

Kinetic energy of a particle executing SHM is K=(1/2)mω 2(a2-y2) =(1/2)× 10( (2π /2) )2[(10)2-(5)2] =375 π 2 texterg

Q. If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position?

$0.375 π^{2} erg$

$3.75 π^{2} erg$

$37.5 π^{2} erg$

$375 π^{2} erg$

Kinetic energy of a particle executing SHM is $K = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$ $= \frac{1}{2} \times 10 (\frac{2 π}{2})^{2} [(10)^{2} - (5)^{2}]$ $= 375 π^{2} erg$

Q. If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position?

0.375π2erg

3.75π2erg

37.5π2erg

375π2erg

Kinetic energy of a particle executing SHM is K=21​mω2(a2−y2) =21​×10(22π​)2[(10)2−(5)2] =375π2erg

$0.375 π^{2} erg$

$3.75 π^{2} erg$

$37.5 π^{2} erg$

$375 π^{2} erg$

Kinetic energy of a particle executing SHM is $K = \frac{1}{2} m ω^{2} (a^{2} - y^{2})$ $= \frac{1}{2} \times 10 (\frac{2 π}{2})^{2} [(10)^{2} - (5)^{2}]$ $= 375 π^{2} erg$