Question

A cheerleader waves her "pom-pom" in SHM with an amplitude of 18.0cm and a frequency of 0.850Hz. Find, a) the maximum magnitudes of acceleration and velocity. b) the acceleration and speed when the pom-pom’s coordinate is x = 9.00cm. c) the time required to move from equilibrium position directly to a point 12.0cm away.

Answer 1

Explanation:

Given that,

Amplitude of wave, A = 18 cm = 0.18 m

Frequency of wave, f = 0.85 Hz

(a) The maximum velocity in SHM is given by :

$v=\omega\times A$

$v=2\pi f\times A$

$v=2\pi \times 0.85 \times 0.18$

v = 0.96 m/s

The maximum acceleration in SHM is given by :

$a=\omega^2\times A$

$a=4\pi ^2f^2\times A$

$a=4\pi ^2\times 0.85^2\times 0.18$

$a=5.13\ m/s^2$

(b) The velocity of particle at point x is given by :

$v=\omega \sqrt{A^2-x^2}$

$v=2\pi f \sqrt{A^2-x^2}$

at x = 9 cm = 0.09 m

$v=2\pi \times 0.85 \sqrt{(0.18)^2-(0.09)^2}$

v = 0.83 m/s

The acceleration of particle at point x is given by :

$a=-\omega^2\times x$

$a=-4\pi ^2f^2\times x$

$a=-4\pi ^2\times 0.85^2\times 0.09$

$a=-2.56\ m/s^2$

(c) The equation of SHM is given by :

$x=Acos(\omega t+\phi)$

Let $\phi=\dfrac{-\pi}{2}$ (at the particle is at x = 0 at t = 0 )

$t=\dfrac{1}{\omega}\ sin^{-1}(\dfrac{x}{A})$

$t=\dfrac{1}{2\pi f}\ sin^{-1}(\dfrac{x}{A})$

$t=\dfrac{1}{2\pi \times 0.85}\ sin^{-1}(\dfrac{0.12}{0.18})$

t = 0.13 seconds

Hence, this is the required solution.