Question

y(t)=25sin(2πt+π/4) magnitude of maximum velocity​

Answer 1

Answer:

50 π m / sec .

Explanation:

Given :

y ( t ) = 25 sin ( 2 π t + π / 4 )

We have standard equation i.e.

y ( t ) = A sin ( ω t + Φ )

On comparing we get :

A = 25 m

ω = 2 π

We know :

v_max =  ω A

v_max =  2 π × 25

v_max =  50 π

Therefore , maximum velocity is  50 π m / sec .

Answer 2

Answer:

• The Maximum velocity of the particle is (v_[max]) = 50 Π m/s.

Given:

1. The given equation y (t) = 25 sin( 2πt + π / 4 ).

Explanation:

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From the standard Equation we Know,

⊕ y (t) = A sin( ωt + ∅ )

where,

• A denotes Amplitude of oscillation.
• ω denotes Angular frequency of the oscillation.
• t denotes time.

Comparing the Given equation with the standard equation we get,

⇒ Amplitude (A) = 25 m.

⇒ Angular Frequency (ω) = 2 π rad/sec.

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Now, Maximum Velocity Formula,

⊕ v_[max] = ω.A

Where,

• ω denotes angular frequency.
• A denotes Amplitude of the SHM.

Substituting the values,

⇒ v_[max] = 2 π x 25

⇒ v_[max] = 50 π m/s.

∴ The Maximum velocity of the particle is (v_[max]) = 50 Π m/s.

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Additional Formulas:

1. v (t) = A cos ( ωt + ∅ )
2. v (t) = ω.(A - y)
3. a (t) = - A cos ( ωt + ∅ )
4. a (t) = ω.(√A² - y²)

Symbols have their usual meaning.

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