Question

The equation of S.H.M. of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase 30° is ____________.

पर्याय

• ${\displaystyle \text{x}=4 \text{sin}(2\pi \text{t}+\pi /6)}$

• ${\displaystyle \text{x}=4 \text{sin}(5\pi \text{t}+\pi /6)}$

• ${\displaystyle \text{x}=4 \text{sin}(2\pi \text{t}+\pi /3)}$

• ${\displaystyle \text{x}=4 \text{sin}(5\pi \text{t}+\pi /3)}$

Answer 1

The equation of S.H.M. of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase 30° is ${\displaystyle \text{x}=4 \text{sin}(5\pi \text{t}+\pi /6)}$.

Explanation:

${\displaystyle \text{A}=4\text{cm},\alpha =30°=\frac{\pi }{6}}$

f = 150 oscillation per minute= 150/60 Hz,

${\displaystyle \omega =2\pi \text{f}=2\pi \times \frac{150}{60}=5\pi }$

Equation of S.H.M becomes,

 ${\displaystyle \text{x}=4 \text{sin}(5\pi \text{t}+\pi /6)}$