A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = A sin (π t/90). The ratio of kinetic to potential energy of this particle at t = 210 s will be :

Question

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = A sin (π t/90). The ratio of kinetic to potential energy of this particle at t = 210 s will be : (A) 2 (B) (1/9) (C)  (1/3) (D) 1

Tardigrade Admin · Accepted Answer

Kinetic Energy, K =(1/2) m ω2 A 2 cos 2 ω t Potential Energy, U=(1/2) m ω2 A2 sin 2 ω t ( K / U )= cot 2 ω t = cot 2 (π/90)(210)= cot 2(2 π+(π/3))=(1/3) Hence ratio is (1/3).

Q. A particle undergoing simple harmonic motion has time dependent displacement given by $x (t) = A sin \frac{π t}{90}$ . The ratio of kinetic to potential energy of this particle at $t = 210 s$ will be :

2

$\frac{1}{9}$

$\frac{1}{3}$

1

Kinetic Energy, $K = \frac{1}{2} m ω^{2} A^{2} cos^{2} ω t$
Potential Energy, $U = \frac{1}{2} m ω^{2} A^{2} sin^{2} ω t$
$< b r / > \frac{K}{U} = cot^{2} ω t = cot^{2} \frac{π}{90} (210) = cot^{2} (2 π + \frac{π}{3}) = \frac{1}{3} < b r / >$
Hence ratio is $\frac{1}{3}$ .

Q. A particle undergoing simple harmonic motion has time dependent displacement given by x(t)=Asin90πt​. The ratio of kinetic to potential energy of this particle at t=210s will be :

2

91​

31​

1

Kinetic Energy, K=21​mω2A2cos2ωt Potential Energy, U=21​mω2A2sin2ωt <br/>UK​=cot2ωt=cot290π​(210)=cot2(2π+3π​)=31​<br/> Hence ratio is 31​.

Q. A particle undergoing simple harmonic motion has time dependent displacement given by $x (t) = A sin \frac{π t}{90}$ . The ratio of kinetic to potential energy of this particle at $t = 210 s$ will be :

$\frac{1}{9}$

$\frac{1}{3}$

Kinetic Energy, $K = \frac{1}{2} m ω^{2} A^{2} cos^{2} ω t$
Potential Energy, $U = \frac{1}{2} m ω^{2} A^{2} sin^{2} ω t$
$< b r / > \frac{K}{U} = cot^{2} ω t = cot^{2} \frac{π}{90} (210) = cot^{2} (2 π + \frac{π}{3}) = \frac{1}{3} < b r / >$
Hence ratio is $\frac{1}{3}$ .