Question

A simple harmonic wave having an amplitude a and time period T
is represented by the equation y = 5 sin r(t + 4)m. Then the
value of amplitude (a) in (m) and time period (T) in second are
(a) a = 10, T = 2
(b) a = 5, T = 1
(©) a = 10, T = 1
(d) a = 5, T = 2
PLZ EXPLAIN it.
who explain properly I mark him as BRAINLIEST ​

Answer 1

Correct Question:-

A simple harmonic wave having an amplitude a and time period T is represented by the equation y = 5 sin π (t + 4) m. Then the value of amplitude (a) in (m) and time period (T) in second are:-

(a) a = 10, T = 2

(b) a = 5, T = 1

(c) a = 10, T = 1

(d) a = 5, T = 2

Answer:

• Amplitude (a) is 5m & Time Period (T) is 2 Sec.

Given:

1. y = 5 sin π(t +4) m [SHM Equation]

Explanation:

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From The Standard SHM Equation.

$\large\bigstar \: {\boxed{\tt y = a sin (\omega t + \phi)}}$

$\bold{Here}\begin{cases}\text{y Denotes Displacement} \\ \text{a Denotes Amplitude} \\ \omega \: \text{Denotes Angular Frequency} \\ \phi \: \text{Denotes Phase}\end{cases}$

$\large{\boxed{\tt y = a sin (\omega t + \phi)}}$

$\large{\tt \hookrightarrow y = a sin (\omega t + \phi)}$

But we Know, ω = 2π/T

Substituting,

$\large{\tt \hookrightarrow y = a sin \Bigg(\dfrac{2 \pi}{T} t + \phi\Bigg)}$

• Here "T" Denotes Time Period.

$\large{\tt \hookrightarrow y = a sin \Bigg(\Bigg[\dfrac{2 \pi}{T}\Bigg] t + \phi\Bigg) \: ----(1)}$

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Now, From the Given Equation.

$\large{\underline{\boxed{\tt y = 5 sin \pi (t + 4)}}}$

$\large{\tt \hookrightarrow y = 5 sin \pi (t + 4)}$

Now,

$\large{\tt \hookrightarrow y = 5 sin (\pi t + 4\pi) \: -----(2)}$

Comparing Equation (1) & (2).

Case-1

Amplitude

By Comparing the equation we get.

$\large{\underline{\boxed{\tt a = 5 \: m}}}$

Case-2

Time period

By Comparing the equation we get.

$\large{\tt \hookrightarrow \Bigg[\dfrac{2 \pi}{T}\Bigg] t = \pi t}$

$\large{\tt \hookrightarrow \Bigg[\dfrac{2 \pi}{T}\Bigg] \cancel{t} = \pi \cancel{t}}$

$\large{\tt \hookrightarrow \Bigg[\dfrac{2 \pi}{T}\Bigg] = \pi }$

$\large{\tt \hookrightarrow \Bigg[\dfrac{2 \cancel{\pi}}{T}\Bigg] = \cancel{\pi}}$

$\large{\tt \hookrightarrow \Bigg[\dfrac{2}{T}\Bigg] = 1 }$

$\large{\tt \hookrightarrow T = 2 \times 1}$

$\large{\underline{\boxed{\tt T = 2 \: Seconds}}}$

∴ Amplitude (a) is 5m & Time Period (T) is 2sec [Option - 4].

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Answer 2

$\Huge{\underline{\underline{\mathfrak{Question \colon }}}}$

A simple harmonic wave having an amplitude a and time period T is represented by the equation y = 5 + sin[π(t + 4) m. Then find the value of amplitude and time period of the wave

(a) a = 10,T = 2

(b) a = 5,T = 1

(c) a = 10,T = 1

(d) a = 5,T = 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Given,

$\large{ \tt{y = 5sin(\pi( t + 4))}.............(1)}$

Standard SHM equation is :

$\large{ \sf{y = a \: sin( \omega \: t \: + \phi) }.....................(2)}$

• a is the amplitude of the wave

On comparing equations (1) and (2),we get :

$\huge{\implies \ \boxed{ \boxed{ \sf{a = 5}}}}$

Thus,the amplitude of the wave is 5

Options (a) and (c) are eliminated

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Now,

$\huge{ \boxed{ \boxed{ \sf{ \omega \: = \frac{2\pi}{T} }}}}$

• T is the time period of the wave

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On comparing (1) and (2) again,we get :

$\large{ \sf{ \frac{2 \cancel{ \pi}}{T} \cancel{t} = \cancel{\pi \: t} }} \\ \\ \large{ \hookrightarrow \: \sf{ \frac{2}{T} = 1}} \\ \\ \huge{ \hookrightarrow \: \boxed{ \boxed{ \sf{T = 2s}}}}$

Thus,the time period of the wave is 2s

Option (b) is eliminated

• Option (d) is the correct option

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