Question

a body oscillates with simple
harmonic motion according to the equation(in SI units) x = 5cos (2πt + π/4). at t = 1.5s calculate i) displacement ii) speed iii) acceleration

Answer 1

(1)
x = 5 cos(2πt + π/4)
At t = 1.5 s
x = 5 cos[(2π × 1.5) + π/4]
x = 5 cos (13π/4)
x = -3.53 m

(2)
v = A ω cos (ωt + φ)
= 5 × (2 π) cos [(2 π × 1.5) + π/4]
[∵ by comparing given equation and general SHM equation ω = 2 π]
= 31.4 × 0.99
= 31.35 m/s

(3)
Acceleration = -A ω² sin(ωt + φ)
= -5 × (2 π)² sin[(2 π × 1.5) + π/4]
= -0.283 rad/s^2

Answer 2

Explanation:

1. x=5cos(2πt+π/4) at t=1.5 s

x=5cos[(2π×1.5)+π/4]

x= 5cos[3π+π/4]

$cos(n\pi + theta ) = { - 1}^{n} \cos(theta)$

x=5(-1)³cosπ/4 ( cosπ/4=45° =1/√2 )

x= -5×1/√2

x=3.53m

2. v=dx / dt

v= d[5cos(2πt+π/4] /dt

v= -5sin(2πt+π/4) . 2π

v= -5×2πsin[2πt+π/4]

v= -5×2π sin [3π+π/4]

$\sin(n\pi + theta) = { -1 }^{n} sin(theta)$

v=-5×2×3.14×(-1)³sinπ/4

v=5 ×2×3.14×1/√2

v=10×3.14/√2

v=22.22m/sec

3. a =dv/dt

a=d[-5×2π sin(2πt+π/4)]

a= -5×2πcos[2πt+π/4].2π at t=1.5 s

a= -5×(2π)²cos 2π×1.5×π/4

a=-5×4π²cos[3π+π/4]

a=-20π²(-1)³cosπ/4

a=20×3.14×3.14/√2

a=197.192/√2

a=139.43m/sec²