Question

consider a shm of time period T. calculate the time takeen for the displacement to change value from half the amplitude to the amplitude
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Answer 1

Answer:

Given:

An SHM of time period T has been provided.

To find:

Time taken to travel from Half-amplitude point to Amplitude.

Concept:

Considering amplitude as "A"

Let the equation of the SHM be as follows:

y = A sin(ωt)

Now , we have to calculate time taken to travel from "A/2" to "A".

This can be found by the following observation :

Time required = {Total time to travel from 0 to A } - { Time taken to travel from 0 to A/2}

Calculation:

Time taken to travel from 0 to A

= T/4 ...........(1)

( As it is ¼ of the total journey)

Time taken to travel from 0 to A/2 can be calculated as

A/2 = A sin(ωt)

=> ½ = sin(ωt)

=> sin(π/6) = sin{(2π/T) × t}

=> π/6 = 2π/T × t

=> t = T/12 .................(2)

So the required time to travel from

A/2 to A

= (1) - (2)

= T/4 - T/12

= T/6

So the answer is T/6.

Answer 2

$\huge \bf{ \red{ \mid{ \overline{ \underline{ANSWER</p><p>}}} \mid}}$

$\huge\bf \purple{ \implies{t = \frac{T}{6}}}$

$\huge \bf \pink{ \mid { \overline{ \underline{GIVEN:-}}} }$

$\sf{ \purple{\implies consider \: the \: time \: period \: t}}$

$\sf \purple{ \implies{and \: amplitude \: a}}$

$\bf\blue{TO \: FIND :-</p><p> }$

→ The time taken for the displacement to change value from half the amplitude to the amplitude

$\large \bf{CALCULATION}$

→ Time period = Value of time at amplitude a - value of time at amplitude a/2.

→ As we know the value of time period at amplitude 0 to a = t/4 ---( i )

$\bf{Equation }$

$\bf{y \: = asin( \omega)t}$

→ At amplitude 0 to a/2

$\bf{ \implies\frac{a}{2} \: = asin( \omega)t} \\ \\ \sf \implies \frac{1}{2} = sin{ (\omega)}t \\ \\ \implies \sf \: \sin \frac{\pi}{6} = sin( \frac{2\pi}{T} )t \\ \\ \sf \implies \:$

$\sf\implies \: \frac{\pi}{6} = \frac{2\pi}{t} \times t \\ \\ \sf \implies \: T\: =\frac{t}{12}..</p><p>.(ii)$

→ Subtract a - a/2

→ t/4 - t/12

→ 2t/12

♦

$\huge \green{\bf{\implies{ \frac{t}{6} }}}$