Question

An object in simple harmonic motion has amplitude 3.2 cmcm and frequency 3.4 HzHz , and at tt = 0ss it passes through the equilibrium point moving to the right. Part A Write the function x(t) that describes the object's position. x(t)=(3.2cm)cos((3.4ra(d)/(s))t) x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t-(\pi )/(2)rad) x(t)=(1.6cm)cos((6.8\pi ra(d)/(s))t) x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t+(\pi )/(2)rad)

Name: an object in simple harmonic motion has amplitude 32 cmcm and frequency 34 hzhz and at tt 0ss it passes through the equilibrium point moving to the right part a write the function xt that de 42236
Uploaded: 2023-04-25T07:59:50-08:00
Duration: 2 min 9 s
Channel: Madhur L
Description: an object in simple harmonic motion has amplitude 32 cmcm and frequency 34 hzhz and at tt 0ss it passes through the equilibrium point moving to the right part a write the function xt that de 42236

          An object in simple harmonic motion has amplitude 3.2 cmcm and frequency 3.4 HzHz , and at tt = 0ss it passes through the equilibrium point moving to the right.
Part A
Write the function x(t) that describes the object's position.
x(t)=(3.2cm)cos((3.4ra(d)/(s))t)
x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t-(\pi )/(2)rad)
x(t)=(1.6cm)cos((6.8\pi ra(d)/(s))t)
x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t+(\pi )/(2)rad)

an object in simple harmonic motion has amplitude 32 cmcm and frequency 34 hzhz and at tt 0ss it passes through the equilibrium point moving to the right part a write the function xt that de 42236

Added by Margarita H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

04/25/2023

Step 1

2 \text{ cm}$ and frequency $3.4 \text{ Hz}$, and at $t = 0 \text{ s}$ it passes through the equilibrium point moving to the right. Write the function $x(t)$ that describes the object's position. $x(t)=(3.2\text{cm}) \cos((3.4 \text{ rad/s})t)$ $x(t)=(3.2\text{cm}) Show more…

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An object in simple harmonic motion has amplitude 3.2 cmcm and frequency 3.4 HzHz , and at tt = 0ss it passes through the equilibrium point moving to the right. Part A Write the function x(t) that describes the object's position. x(t)=(3.2cm)cos((3.4ra(d)/(s))t) x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t-(\pi )/(2)rad) x(t)=(1.6cm)cos((6.8\pi ra(d)/(s))t) x(t)=(3.2cm)cos((6.8\pi ra(d)/(s))t+(\pi )/(2)rad)