Question

A body describes shm with amplitude of 5 cm and a period of 0.2seconds. Find the acceleration and velocity of the body

Answer 1

Amplitude ( A ) = 5 cm

Time period ( T) = 0.2 sec

(a) when displacement ( x) is 5 cm

Then,

a = -w²x

= -(2π/T)²x

= - (2π/0.2)² × 5 × 10^-2

= -5π² m/s

Velocity = w√(A² - x²)

= (2π/T)√(A² - x²)

= (2π/0.2)√(0.05² - 0.05²)

= 0

(b) when x = 3 cm ,

acceleration = -w²x

= ( 2π/0.2)² × 3 × 10^-2

= (10π)² × 3 × 10^-2

= 3π² m/s

Velocity = w√(A² - x²)

= (2π/0.2)√(0.05²-0.03²)

= 0.4π² m/s

(c) when displacement is x = 0

Acceleration a = - w²x = 0

Velocity = w√(A² - x²) = wA

= (2π/0.2)√0.05²

= 0.5π m/s

hope it will help you

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Answer 2

Given :

▪ A body is performing SHM with amplitude of 5cm.

To Find :

▪ Max. velocity and Max. acceleration of body.

Formula :

✒ Maximum velocity :

$\bigstar\:\underline{\boxed{\bf{\pink{v_{max}=A\omega}}}}$

✒ Maximum acceleration :

$\bigstar\:\underline{\boxed{\bf{\green{a_{max}=A{\omega}^2}}}}$

• A denotes amplitude
• ω denotes angular frequency

✒ Formula of angular frequency in terms time period is given by

$\bigstar\:\underline{\boxed{\bf{\orange{\omega=\dfrac{2\pi}{T}}}}}$

• T denotes time period

____________________________

✴ Angular frequency :

✏ ω = 2π / T

✏ ω = 2×π / 0.2

✏ ω = 10π Hz

✴ Maximum velocity :

→ Vmax = Aω

→ Vmax = 0.05×10π

→ Vmax = 0.5π mps

✴ Maximum acceleration :

✏ (a)max = Aω^2

✏ (a)max = 0.05×(10π)^2

✏ (a)max = 5π^2 m/s^2