Question

a particle executes simple harmonic motion of period 10 seconds and amplitude of 1.5 metre. calculate its maximum velocity and acceleration​

Answer 1

Explanation:

T= 10 sec

A= 1.5 m

Vmax= ?

amax= ?

Vmax= WA

=2π/T × A = 2 × 3.14/10 × 1.5

=0.94 m/s

amax= W²A

=2π/T × A = (2 × 3.14/10)² × 1.5

=0.59 m/s²

Answer 2

Maximum velocity is 0.94 m/s and maximum acceleration is 1.413 m/s².

Given:

A particle executes simple harmonic motion of period 10 seconds and amplitude of 1.5 metre.

To find:

Maximum velocity and maximum acceleration

Solution:

T = 10s and A = 1.5m

Vmax = ωA = $\frac{2\pi }{T} *A$ = $\frac{2 * 3.14}{10} * 1.5$

⇒ Vmax = 0.94 m/s

Amax = ωA² = $\frac{2\pi }{T} *A^{2}$ = $\frac{2 * 3.14}{10} * 1.5^{2}$

⇒ Amax = 1.413 m/s²

Therefore maximum velocity is 0.94 m/s and maximum acceleration is 1.413 m/s².

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