Answer:
d=[tex]5[/tex]
Step-by-step explanation:
The given simple harmonic motion equation is given as:
d=[tex]5sin(2{\pi}t)[/tex]
Differentiating the above equation with respect to t, we have
[tex]d'=5cos(2{\pi}t)(2{\pi)[/tex]
[tex]d'=10{\pi}cos(2{\pi}t)[/tex]
Now, equating [tex]d'=0[/tex], we have
[tex]10{\pi}cos2{\pi}t=0[/tex]
[tex]cos2{\pi}t=0[/tex]
[tex]2{\pi}t=\frac{{\pi}}{2}[/tex]
[tex]t=\frac{1}{4}[/tex]
Also, differentiating again with respect to t, we get
[tex]d''=10{\pi}(-sin(2{\pi}t)(2{\pi})[/tex]
⇒[tex]d''<0[/tex]
Therefore, at t=[tex]\frac{1}{4}[/tex], d=[tex]5sin(2{\pi}t)[/tex] will have maximum displacement.
⇒d=[tex]5sin(2{\pi}(\frac{1}{4}))[/tex]
⇒d=[tex]5sin(\frac{\pi}{2})[/tex]
⇒d=[tex]5[/tex]
