Question

if v is the maximum speed of a particle in SHM of amplitude a its average speed in one time period is​

Answer 1

Answer:

Given:

Maximum velocity of particle in SHM of amplitude "a" is "v"

To find:

In one time period, the average velocity of the object.

Concept:

Always remember that Average velocity is the ratio of total distance to the total time taken.

Calculation:

We know that :

$v \: max. =( a \times w) =v$

ω => angular frequency

a => amplitude

Now ,

$v \: avg. = \frac{distance}{time}$

$= > v \: avg. = \frac{4a}{t}$

$= > v \: avg. = \frac{4a}{( \frac{2\pi}{w}) }$

$= > v \: avg. = \frac{4aw}{ 2\pi}$

$= > v \: avg. = \frac{4}{ 2\pi} \times (aw)$

$= > v \: avg. = \frac{4}{ 2\pi} \times (v )$

So, final answer is (4v/2π).

Answer 2

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

We know that :

$\Large{\underline{\boxed{\sf{v_{avg} \: = \: \dfrac{Displacement}{time}}}}}$

Here Displacement = 4a

Putting values

$\rightarrow {\sf{v_{avg} \: = \: \dfrac{4a}{\dfrac{2 \pi}{\omega}} \: \: \: \: \: \big[ \because \: \omega \: = \: \dfrac{2 \pi}{t} \big]}}$

$\rightarrow {\sf{v_{avg} \: = \: \dfrac{4 \omega}{2 \pi}}}$

$\rightarrow {\sf{v_{avg} \: = \: \dfrac{4 (a \: \times \: \omega)}{2 \pi}}}$

$\rightarrow {\sf{v_{avg} \: = \: \dfrac{4v}{2 \pi} \: \: \: \: \: [\because \: v_{max} \: = \: a \: \times \: \omega]}}$

$\large \implies {\boxed{\sf{v_{avg} \: = \: \dfrac{4v}{2 \pi}}}}$