Question

A pendulum is swinging back and forth. The angular frequency of the oscillation is = radians/s, and the phase shift is = 0 radians. At time t = 8.50 s, the pendulum is 14.0 cm from its equilibrium position. What is the amplitude of the oscillation?

• A. 14 cm
• B. 15 cm
• C. 16 cm
• D. 18 cm

Answer 1

The correct option is A 14 cm

The position of the pendulam at a given time is the variable $x$ which has a value $x=14cm$ or $x=0.14m$. The amplitude $A$ can be found by rearranging the formula

$x=A\sin (wt+\varphi )⟶A=\frac{x}{\sin (wt+\varphi )}$

$A=\frac{x}{\sin (wt+\varphi )}$

$A=\frac{0.14m}{\sin [\left(\pi rad/s\right)\left(8.505\right)+0]}$

$A=\frac{0.140m}{\sin \left(8.50\pi \right)}$

A sine of $8.50\pi$ can be solved

i.e. $\sin \left(8.50\pi \right)=1$

Therefore $A=\frac{0.140m}{\sin \left(8.50\pi \right)}=\frac{0.140}{1}=0.140m$

The amplitude of the perpendicular ossicilators is $A=0.140m=14cm$