Question

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Answer 1

Given:

A body of 0.25 kg executes SHM given by

$y = 0.4 \sin(0.5\pi t)$

To find:

• Amplitude
• Angular frequency
• maximum Velocity
• maximum acceleration

Calculation:

General Equation of SHM is :

$y = a \sin( \omega t)$

• Here 'a' is amplitude.

So, comparing this with our equation, we get :

• Amplitude is 0.4 metres.

Again, angular frequency $\omega$:

$\omega = 0.5\pi = 1.57 \: hz$

Now, Velocity can be calculated as:

$v = \dfrac{dy}{dt}$

$\implies \: v = \dfrac{d \{0.4 \sin(0.5\pi t) \}}{dt}$

$\implies \: v = 0.4 \times0.5\pi \times \cos(0.5\pi t)$

• Max value of sin is 1

$\implies \: v_{max} = 0.4 \times0.5\pi$

$\implies \: v_{max} = 0.62 \: m {s}^{ - 1}$

So, max Velocity is 0.62 m/s.

Max acceleration is :

$| acc_{max} | = { \omega}^{2} a$

$\implies | acc_{max} | = {(0.5\pi)}^{2} \times 0.4$

$\implies | acc_{max} | = 0.98 \: m {s}^{ - 2}$

So , max acceleration is 0.98 m/s².