a particle moves along a straight line according to the law x = a cos(nt +k). prove that its acceleration is directed towards the origin and varies as the distance

Open in App

Solution

Dear Student,
As given the displacement, $x = a \cos (n t + k)$
So, the velocity,
$v = \frac{d x}{d t} = \frac{d}{d t} (a \cos (n t + k)) = a [- \sin (n t + k) . \frac{d}{d t} (n t + k) = - n . a \sin (n t + k)$
Now, acceleration , $a = \frac{d v}{d t} = \frac{d}{d t} [- n . a \sin (n t + k)] = - n^{2} a \cos (n t + k) = - n^{2} x [a s g i v e n i n t h e q u e s t i o n, x = a \cos (n t + k)] S o w e c a n s a y t h a t, a c c e l e r a t i o n i s d i r e c t e d t o w a r d s t h e o r i g i n a n d v a r i e s a s t h e d i s \tan c e$

Regards

Suggest Corrections

Similar questions

k \sqrt{a^{2} - x^{2}}

v^{2} = 4 a (x sin x + cos x)

v

x

x^{2} = α t^{2} + 2 β t + γ

x

α, β, γ

x = \frac{1}{2} v t, x

t

v

s = 16 - 2 t + 3 t^{3}

s

t

2

Join BYJU'S Learning Program