Question

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion 10 $S^{-1}$ . At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is $\frac{\pi }{4}$ .

• A. $750\sqrt{2}m/s^2$
• B. $500\sqrt{2}m/s^2$
• C. $750m/s^2$
• D. $500m/s^2$

Answer 1

Answer: (B)

$500\sqrt{2}m/s^2$

$a_{max}=A{\omega }^2$ .....(1)

$V_{max}=A\omega$

Now,

$\frac{a_{max}}{V_{max}}=\frac{A{\omega }^2}{A\omega }=\omega$

i.e $=\omega ={10}^o$

displacement $=A_{m}cos(\omega t+\theta )$

$=Acos(0+\frac{1}{4})$

$=Acos\frac{1}{4}$

$5=A\frac{1}{\sqrt{2}}$

$\Rightarrow A=5\sqrt{2}$

Now $a_{max}=5\sqrt{2}\times {10}^2$ from (1)

$=10\times 5\sqrt{2}$

$=500\sqrt{2}m/s^2$