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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

If the displacement $x$ and velocity $v$ of a particle executing simple harmonic motion are related through the expression $4 v^{2} = 25 - x^{2}$ , then its time period is

Q. The displacement of a particle executing S.H.M is given by

x

0.25 s i n 200 t

in meters, the maximum speed of particle is:

Q. A particle is executing S.H.M. with amplitude

^{'} a^{'}

and has maximum velocity

^{'} v^{'}

. Its speed at displacement

a / 2

will be

The velocity v and displacement x of a particle executing simple harmonic motion are related as

$v \frac{d v}{d x} = - ω^{2} x$

At $x = 0, v = v_{0}$ . Find the velocity v when the displacement becomes x.

Join BYJU'S Learning Program

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 is

The correct option is B 5v=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

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