Question

a needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. find its acceleration (1/30)s after it has crossed the mean position?
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Answer 1

Given :

Amplitude = 4 cm = 0.04 m

Frequency = 5 Hz

To Find :

Acceleration of needle after (1/30) second after it has crossed mean position.

Solution :

❖ In such questions first of all check that what is asked to find. Here we are asked to find acceleration of the needle. Now just note down formula of acceleration.

» Acceleration of particle at a distance of x from the mean position is given by

$\dag\:\underline{\boxed{\bf{\orange{a=-\omega^2x}}}}$

• Negative sign shows that it acts towards mean position always. [opposite to the direction of displacement]

Now we have to find values of angular velocity (ω) and displacement (x).

♦ Angular velocity of needle :

➙ ω = 2 π f

where f denotes frequency

➙ ω = 2 π (5)

➙ ω = 10 π rad s‾¹

♦ Displacement of needle :

Displacement of an oscillating particle from the mean position after time t in terms of amplitude and time is given by

➙ x = A sin (ω t)

➙ x = 0.04 sin (10π × 1/30)

➙ x = 0.04 sin (π/3)

• sin (π/3) = √3/2

➙ x = 0.04 × √3/2

➙ x = 0.0346 m

By substituting the values in formula;

➠ a = ω² x

➠ a = (10 π)² × 0.0346

➠ a = 3.46 × (3.14)²

➠ a = 34.11 m/s² [towards the mean position]

Answer 2

$\begin{gathered} \\ \Large{\bf{\red{\underline{Given\::}}}} \\ \end{gathered}$

Amplitude = 4 cm = 0.04 m

Frequency = 5 Hz

$\begin{gathered} \\ \Large{\bf{\green{\underline{To \: Find:}}}} \\ \end{gathered}$

Acceleration of needle after (1/30) second after it has crossed mean position.

$\begin{gathered} \\ \Large{\bf{\red{\underline{Solution \::}}}} \\ \end{gathered}$

❖ In such questions first of all check that what is asked to find. Here we are asked to find acceleration of the needle. Now just note down formula of acceleration.

» Acceleration of particle at a distance of x from the mean position is given by

by

a=−ω²x

• Negative sign shows that it acts towards mean position always. [opposite to the direction of displacement]

♦ Angular velocity of needle :

➙ ω = 2 π f

where f denotes frequency

➙ ω = 2 π (5)

➙ ω = 10 π rad s‾¹

♦ Displacement of needle :

Displacement of an oscillating particle from the mean position after time t in terms of amplitude and time is given by

➙ x = A sin (ω t)

➙ x = 0.04 sin (10π × 1/30)

➙ x = 0.04 sin (π/3)

sin (π/3) = √3/2

➙ x = 0.04 × √3/2

➙ x = 0.0346 m

By substituting the values in formula;

➠ a = ω² x

➠ a = (10 π)² × 0.0346

➠ a = 3.46 × (3.14)²

➠ a = 34.11 m/s² [towards the mean position]