Respuesta :
Answer:
A) Period of the motion:**
- The period (T) is the time it takes for the object to complete one full oscillation.
- It's related to the angular frequency (ω) by the formula: T = 2π/ω
- In this case, ω = 8.0π (from the given expression), so T = 2π/(8.0π) = **0.25 seconds
(b) Frequency of the motion:**
- The frequency (f) is the number of oscillations per second.
- It's the reciprocal of the period: f = 1/T = 1/0.25 = **4.0 Hz**.
(c) Amplitude of the motion:**
- The amplitude (A) is the maximum displacement of the object from its equilibrium position.
- It's given directly in the expression as 5.2 cm.
(d) First time after t = 0 that the object reaches x = 2.6 cm:**
- We need to solve the equation x = 2.6 cm = 5.2 cm sin(8.0πt) for t.
- Divide both sides by 5.2 cm: sin(8.0πt) = 0.5.
- Find the arc sine of both sides: 8.0πt = arcsin(0.5).
- Divide both sides by 8.0π: t = arcsin(0.5) / (8.0π).
- Using a calculator, we get t ≈ **0.013 seconds
Answer:
a) Period: [tex]\sf 0.25[/tex] s
b) Frequency: [tex]\sf 4[/tex] Hz
c) Amplitude: [tex]\sf 5.2[/tex] cm
d) First time reaching [tex]\sf x = 2.6[/tex] cm: [tex]\sf 0.0208[/tex] s
Explanation:
Simple Harmonic Motion (SHM) :Back-and-forth wiggle, pull brings it back, like a spring or a pendulum's swing.
Let's find the answer for given questions:
a) Period:
The period [tex]\sf (T) [/tex] of the motion is calculated using the formula [tex]\sf T = \dfrac{2\pi}{\omega}[/tex], where [tex]\sf \omega[/tex] is the angular frequency.
In this case:
First let's compare the given expression [tex]\sf x=(5.2cm)\, sin(8.0\pi t) [/tex] with [tex]\sf x = A sin(\omega t)[/tex]
We get
- A=5.2 cm
- [tex]\sf \omega=8\pi \textsf{rad/s}[/tex]
We have, formula to calculate period is:
[tex]\sf T = \dfrac{2\pi}{\omega}[/tex]
Now, substitute the value:
[tex]\sf \begin{aligned} \textsf{Time Period(T)} &= \dfrac{2\pi}{8\pi \textsf{ rad/s} \\\\ &=\dfrac{1}{4} \\\\ &= 0.25 \, \textsf{s} \end{aligned} [/tex]
b) Frequency:
The frequency [tex]\sf (f)[/tex] is the reciprocal of the time period:
Mathematically:
[tex]\sf f = \dfrac{1}{T}[/tex]
Therefore:
[tex]\sf \begin{aligned} \textsf{Frequency (f)} &= \dfrac{1}{0.25} \, \textsf{Hz} \\\\ &= 4 \, \textsf{Hz} \end{aligned}[/tex]
c) Amplitude:
The amplitude of the motion is the absolute value of the coefficient of the sine term.
In this case:
[tex]\sf \textsf{Amplitude} = |5.2 \, \textsf{cm}| \\\\ = 5.2 \, \textsf{cm} [/tex]
d) First time reaching [tex]\sf x = 2.6[/tex] cm:
To find the first time after [tex]\sf t = 0[/tex] when [tex]\sf x = 2.6[/tex] cm, we set up the equation:
[tex]\sf \begin{aligned} x \, \textsf{cm} &= (5.2 \, \textsf{cm})\sin(8.0\pi t) \\\\ 2.6 \, \textsf{cm} &= (5.2 \, \textsf{cm}) \sin(8.0\pi t) \\\\ \sin(8.0 \pi t) & = \dfrac{2.6}{5.2} \\\\ \sin(8.0 \pi t) & = 0.5 \\\\ 8.0 \pi t &= sin^{-1}(0.5) \\\\ 8.0 \pi t & = \dfrac{\pi}{6} \\\\ t & = \dfrac{\pi }{6 \cdot 8.0\pi } \\\\ t & = \dfrac{ 1}{48} \\\\ t & = 0.02083333333 \\\\ t & = 0.0208 \textsf{s (in 4 d.p.} \end{aligned}[/tex]
Therefore, the summarized answers are:
- a) Period: [tex]\sf 0.25[/tex] s
- b) Frequency: [tex]\sf 4[/tex] Hz
- c) Amplitude: [tex]\sf 5.2[/tex] cm
- d) First time reaching [tex]\sf x = 2.6[/tex] cm: [tex]\sf 0.0208[/tex] s
