If displacement x and velocity v are related as 4 v2=25-x2 in a SHM Then time period of given SHM is (consider SI units)

Question

If displacement x and velocity v are related as 4 v2=25-x2 in a SHM Then time period of given SHM is (consider SI units) (A) π (B) 2 π (C) 4 π (D) 6 π

Tardigrade Admin · Accepted Answer

Given 4 v2=25-x2 Differentiating with respect to t on both sides, 8 v a=-2 x v ⇒ a=-(x/4) Motion is simple harmonic: So, ω2=(1/4) and T=2 π √4=4 π

Q. If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM Then time period of given SHM is (consider SI units)

$π$

$2 π$

$4 π$

$6 π$

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 xv \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π 4 = 4 π$

Q. If displacement x and velocity v are related as 4v2=25−x2 in a SHM Then time period of given SHM is (consider SI units)

π

2π

4π

6π

Given 4v2=25−x2 Differentiating with respect to t on both sides, 8va=−2xv⇒a=−4x​ Motion is simple harmonic: So, ω2=41​ and T=2π4​=4π

Q. If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM Then time period of given SHM is (consider SI units)

$π$

$2 π$

$4 π$

$6 π$

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 xv \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π 4 = 4 π$