Question

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

Answer 1

Answer:

5m, 2 sec

Explanation:

y = A sin ( wt + ¢ ) ( general equation)

y = 5 sinπ (t + 4) ( given equation )

re-writng the above.....

y = 5 sin ( πt + 4π )

on comparing....

A = 5m & w = π

but....

w = 2π/ T

hence....

2π/ T = π

T = 2 sec

Answer 2

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\bf{Amplitude,A = 5\:m}}$

• $\boxed{\bf{Time\:Period, T = 2\: sec}}$

⍟ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ⍟

The given equation of motion is

$\:\:\:\: y = 5 sin[\pi (t + 4)]\:\:m$

Standard equation of a particle in SHM is

$\:\:\:\: y = A sin(\omega t + \phi)$

Comparing the equation, we get

Amplitude, $A = 5\:m$

Angular Frequency, $\omega = \pi\:\:rad/s$

So, Time Period(T) is given by,

$\:\:\:\:\: T = \frac{2\pi } {\omega}$

$\to T = \frac{2\pi } {\pi}$

$\to T = 2\: sec$