Question

In an electric shaver, the blade moves back and forth over a distance of 2mm .The motion is SHM with frequency 120 hz find maximum speed and acceleration of blades . ​

Answer 1

Question :

In an electric shaver, the blade moves back and forth over a distance of 2mm .The motion is SHM with frequency 120 hz find maximum speed and acceleration of blades .

Solution :

From the Question,

• Total Displacement,D = 2 mm

• Frequency,f = 120 Hz

Firstly,we need to calculate the amplitude of the blade

Amplitude is the maximum displacement from the mean position. Here,it would be half of the Total Displacement.

• Amplitude,A = 1 mm

$\longrightarrow \: \sf{A = {10}^{ - 3} \: mm}$

Relationship between angular velocity and frequency,

$\boxed{ \boxed{ \sf{ \omega = 2\pi \: f}}}$

Maximum Velocity

$\sf{ v = \omega \: A } \\ \\ \longrightarrow \: \sf{v = (2\pi \: f) \times A} \\ \\ \longrightarrow \: \sf{v = 2 \times 3.14 \times 120 \times {10}^{ - 3} } \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf{v = 0.75 \: {ms}^{ - 1} }}}$

Maximum Acceleration

$\sf{a = { \omega}^{2} A} \\ \\ \longrightarrow \: \sf{a = (2\pi \: f) {}^{2} \times A} \\ \\ \longrightarrow \: \sf{a = 4 \times (3.14) {}^{2} \times {120}^{2} \times {10}^{ - 3} } \\ \\ \longrightarrow \: \boxed{ \boxed{\sf{a = 5.7 \times {10}^{2} \: {ms}^{ - 2} }}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Frequency (v) = 120 Hz
• Distance = 2 mm = 2 × 10^(-3) m
• Amplitude (A) = 1 × 10^(-3) m

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

We have formula :

$\large \star {\boxed{\sf{V \: = \: A \: \omega}}} \\ \\ : \implies {\sf{V \: = \: A \: \times \: \dfrac{2 \pi}{T}}} \\ \\ \small : \implies {\sf{V \: = \: A \: 2 \pi \nu}} \\ \\ \small : \implies {\sf{V \: = \: 1 \: \times \: 10^{-3} \: \times \: 2 \: \times \: 3.14 \: \times \: 120}} \\ \\ \small : \implies {\sf{V \: = \: 0.75}} \\ \\ \large {\boxed{\sf{Max. \: Velocity \: = \: 0.75 \: ms^{-1}}}}$

$\rule{200}{2}$

And formula for Max. Acceleration is :

$\large \star {\boxed{\sf{a \: = \: \omega^2 A}}} \\ \\ : \implies {\sf{a \: = \: (2 \pi \nu)^2 \: \times \:A}} \\ \\ \small : \implies {\sf{a \: = \: 2 \: \times \: (3.14)^2 \: \times \: (120)^2 \: \times \: 1 \: \times \: 10^{-3}}} \\ \\ \small : \implies {\sf{a \: = \: 5.7 \: \times \: 10^{2}}} \\ \\ \large {\boxed{\sf{Acceleration \: = \: 5.7 \: times \: 10^{2} \: ms^{-2}}}}$