A particle intially at the mean position is executing simple harmonic motion with an angular frequency π4 rad/s. The ratio of the distances travelled by the particle in the first second & second is

At t=1 secW=π4rad/sx1=A sin (π4×1) x1=A2 -----------(1)Distance traveled in 2 second t=2secx2=A sin[π4(2)]=A sinπ2=ADistance travelled in 2nd Second x2−x1=A−A2=A[1−12]−−−−−−(2)A2A[1−12]=2+11

A particle intially at the mean position is executing simple harmonic motion with an angular frequency π4 rad/s. The ratio of the distances travelled by the particle in the first second & second is

At t=1 secW=π4rad/sx1=A sin (π4×1) x1=A2 -----------(1)Distance traveled in 2 second t=2secx2=A sin[π4(2)]=A sinπ2=ADistance travelled in 2nd Second x2−x1=A−A2=A[1−12]−−−−−−(2)A2A[1−12]=2+11

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A particle intially at the mean position is executing simple harmonic motion with an angular frequency $\frac{π}{4}$ rad/s. The ratio of the distances travelled by the particle in the first second & second is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1:1

$1 : \sqrt{2} - 1$

$(1 + \sqrt{3}) : 1$

2:1

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At t=1 sec
$W = \frac{π}{4} r a d / s$
$x_{1} = A \sin (\frac{π}{4} \times 1)$
$x_{1} = \frac{A}{\sqrt{2}}$ -----------(1)
Distance traveled in 2 second
$t = 2 \sec$
$x_{2} = A \sin [\frac{π}{4} (2)] = A \sin \frac{π}{2} = A$
Distance travelled in $2^{n d}$ Second
$x_{2} - x_{1} = A - \frac{A}{\sqrt{2}} = A [1 - \frac{1}{\sqrt{2}}] - - - - - - (2)$
$\frac{A \sqrt{2}}{A [1 - \frac{1}{\sqrt{2}}]} = \frac{\sqrt{2} + 1}{1}$