The displacement of a particle performing simple harmonic motion is given by, x=8sinω t+6cos ω t, where distance is in cm and time is in second. The amplitude of motion is

Question

The displacement of a particle performing simple harmonic motion is given by, x=8sinω t+6cos ω t, where distance is in cm and time is in second. The amplitude of motion is (A) 10 cm (B) 2 cm (C) 14 cm (D) 3.5 cm

Tardigrade Admin · Accepted Answer

Here x=8 sin ω t+6 cos ω t so, a1 =8cm and a2 =6cm ∴ Amplitude of motion A =√a12 +a22 =√82+62 =√64+36=√100 =10 cm

Q. The displacement of a particle performing simple
harmonic motion is given by, x=8sin $ω t + 6 cos ω t$ ,
where distance is in cm and time is in second. The
amplitude of motion is

10 cm

2 cm

14 cm

3.5 cm

Here $x = 8 s inω t + 6 cos ω t$
so, $a_{1} = 8 c m an d a_{2} = 6 c m$
$∴$ Amplitude of motion
$A = a_{1}^{2} + a_{2}^{2}$
$= 8^{2} + 6^{2}$
$= 64 + 36 = 100$
$= 10 c m$

Q. The displacement of a particle performing simple harmonic motion is given by, x=8sinωt+6cosωt, where distance is in cm and time is in second. The amplitude of motion is