Question

a body oscillates with SHM according to the equation x=5cos[(2πt)+π/4]. At t=1.5, its acceleration is ​

Answer 1

Answer:

Given:

Equation of SHM has been provided.

x = 5 cos[(2πt) + π/4]

To find:

Acceleration at t = 1.5 seconds.

Concept:

Acceleration is obtained by 2nd order Differentiation of the displacement function wrt time.

$\boxed{ \huge{ \red{acc. = \dfrac{ {d}^{2}x }{ d {t}^{2} }}}}$

Calculation:

x = 5 cos[(2πt) + π/4]

∴ v = dx/dt

=> v = 5 × (-2π) sin[(2πt) + π/4]

=> v = (-10π) sin[(2πt) + π/4]

Now acc = dv/dt = d²x/dt²

=> acc. = (-10π)×(2π) cos[(2πt) + π/4]

=> acc. = (-20π²) cos[(2πt) + π/4]

Now putting value of t = 1.5 secs.

we get :

∴ acc. = (-20π²) cos[{2π(3/2)} + π/4]

=> acc. = (-20π²) cos[3π + π/4]

=> acc. = (-20π²) cos[13π/4]

Value of cos(13π/4) = (-1/√2)

=> acc. = (-20π²) × (-1/√2)

=> acc. = (20√2π²)/2

=> acc. = 10√2 π²

=> acc. = 10 × 1.414 × π × π

=> acc. = 139.57728 m/s²

So final answer :

$\boxed{\red{acc. \:= \: 139.57 \: ms^{-2}}}$

Answer 2

Question :

A body oscillates in SHM according to the equation. At T = 1.5s,the acceleration is__

$\sf \: x = 5 \cos(2\pi \: T + \dfrac{\pi}{4} )$

Solution :

The SHM is represented by the equation :

$\sf \: x = 5 \cos(2\pi \: T + \dfrac{\pi}{4} )$

To finD

Acceleration of the body in SHM

$\rule{300}{2}$

Differentiating x w.r.t t,we get velocity :

$\sf \: v = \dfrac{dx}{dt} \\ \\ \longmapsto \: \boxed{\sf \: v = - 10\pi \: \sin(2\pi \: T + \dfrac{\pi}{4} ) }$

$\rule{300}{2}$

Differentiating v w.rt t,we get acceleration :

$\sf \: a = \dfrac{dv}{dt} \\ \\ \longmapsto \: \boxed{ \sf \: a = - 20 {\pi}^{2} \cos(2\pi \: T + \dfrac{\pi}{4} ) }$

Putting T = 1.5s,

$\longmapsto \: \sf \: a = - 20 \times 3.14 {}^{2} \cos( \frac{13\pi}{4} ) \\ \\ \longmapsto \: \sf \: a = - 20 \times 9.8 \times - 0.707 \\ \\ \longmapsto \: \boxed {\boxed{ \sf \: a = 139.41 \: {ms}^{ - 2}}}\ \ \ \sf[Approx.]$