Question

A simple harmonic motion having an amplitude A and time period T is represented by the equation: y=5sin2π(t+4) m. Then the values of A (in m) and T (in sec) are:​

Answer 1

Answer:

Comparing the equation of SHM with the standard equation,

The amplitude of the oscillation will be a=5 m.

The angular frequency of the oscillation will be ω=π

So, time period will be:

T=  

ω

2π

​  

 

T−  

π

2π

​  

 

=2 sec

Explanation:

Answer 2

Standard equation of SHM,

$y = a\times sin(\omega t+\phi)$

Comparing that with the given equation; $y = 5\times sin\pi(t+ 4)$, we get,

$Amplitude, a = 5 \,m\\\\Angular Frequency, \,\omega= \pi\\\\So, \,\,Time\,\,Period,\\T = \frac{2\pi}{\omega}\\\implies T = \frac{2\pi}{\pi}\\\therefore T = 2\,sec$

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