#### To determine

**a) $2:** the displacement function y=8e−t/2sin4t together with y=8e−t/2 and y=−8e−t/2

#### Answer

y=−8e−t/2 is a reflection of y=8e−t/2 along X axis.

#### Explanation

The graph of the following functions y=8e−t/2sin4t,y=8e−t/2 and y=−8e−t/2 is given below

From the graph of y=8e−t/2, we can see that the graph of y=−8e−t/2 is a reflection of the the graph y=8e−t/2 along X axis.

**Conclusion:** y=−8e−t/2 is a reflection of y=8e−t/2 along X axis.

#### To determine

**b) $2:** the maximum value of the displacement y=8e−t/2sin4t

#### Answer

t=0.3616

#### Explanation

The maximum value of the displacement y=8e−t/2sin4t occurs when it touches y=8e−2t i.e. t=0.3616

**Conclusion:** t=0.3616

#### To determine

**c) $2:** the velocity of the object when it first returns to its equilibrium positions

#### Answer

dydt|t=0.3616=16.709

#### Explanation

Given the displacement y=8e−t/2sin4t, the velcocity is given by dydt

dydt=8[4e−t/2cos4t−12e−t/2sin4t]=4e−t/2[8cos4t−sin4t]

dydt=0⇒8cos4t−sin4t=0⇒tan4t=8⇒t=14tan−1(8)

Velocity of the object when it first returns to its equilibrium position at t=0.3616 is given by dydt|t=0.3616=16.709

**Conclusion:** dydt|t=0.3616=16.709

#### To determine

**d) $2:** the graph and estimate the time after which displacement is not more than 2 cm

#### Answer

**Solution**: See the graph t=2.15

#### Explanation

From the graph the displacement is not more than 2 cm, when time is 2.15

**Conclusion: **t=2.15