Physics - SHM Question with Solution | TestHub

Comparing the given equation $Y=3\sin 2\pi \left(\frac{t}{0.04}-\frac{x}{0.01}\right)$ with the standard wave equation $Y=A\sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$, we get:

Amplitude $A=3$ cm

Time period $T=0.04$ s

Frequency $f=\frac{1}{T}=\frac{1}{0.04}=25$ Hz

Angular frequency $\omega =2\pi f=2\pi \times 25=50\pi$ rad/s

Maximum acceleration $a_{max}={\omega }^{2}A$

$a_{max}=(50\pi {)}^2\times 3=2500{\pi }^2\times 3=7500{\pi }^2$

Given ${\pi }^2=10$, we have:

$a_{max}=7500\times 10=75000$ cm/s${}^2=7.5\times 10^4$ cm/s${}^2$

Answer: $25\mathrm{Hz},7.5\times 10^4\mathrm{cm}/{\mathrm{s}}^2$