If the displacement x and velocity v of a particle executing S-H-M are related through the expression 4v-2-25-x-2- - then its maximum displacement in meters is2516

Question

If the displacement x and velocity v of a particle executing S-H-M are related through the expression  4v-2-25-x-2- - then its maximum displacement in meters is2516

Toppr Admin · Accepted Answer

V-x3C9-x221A-A2-x2212-x24v2-52-x2212-x2-x27F9-v-12-x221A-52-x2212-x2-Comparing the equation with the standard equation we get A-5m

If the displacement $x$ and velocity $v$ of a particle executing S.H.M are related through the expression $4 v^{2} = 25 - x^{2}$ , then its maximum displacement in meters is
2
5
1
6

$v = ω \sqrt{A^{2} - x^{2}}$
$4 v^{2} = 5^{2} - x^{2} ⟹ v = \frac{1}{2} \sqrt{(5^{2} - x^{2})}$
Comparing the equation with the standard equation we get $A =$ 5m

If the displacement x and velocity v of a particle executing S.H.M are related through the expression 4v2=25−x2, then its maximum displacement in meters is2516

v=ω√A2−x24v2=52−x2⟹v=12√(52−x2)Comparing the equation with the standard equation we get A=5m

If the displacement $x$ and velocity $v$ of a particle executing S.H.M are related through the expression $4 v^{2} = 25 - x^{2}$ , then its maximum displacement in meters is
2
5
1
6

$v = ω \sqrt{A^{2} - x^{2}}$
$4 v^{2} = 5^{2} - x^{2} ⟹ v = \frac{1}{2} \sqrt{(5^{2} - x^{2})}$
Comparing the equation with the standard equation we get $A =$ 5m