\[ v_x(t) = ( 2.60 cm/s) \sin \big[ ( 4.63 rad/s ) t – (\pi/2) \big] \]
- The Period
- The Amplitude
- Maximum Acceleration of the Mass
- Force Constant of the Spring
The question aims to find the period, amplitude, acceleration, and force constant of the spring of a mass attached to a spring.
The question is based on the concept of simple harmonic motion (SHM). It is defined as a periodic motion of a pendulum or a mass on a spring. When it moves to and fro is called simple harmonic motion. The equation of the velocity is given as:
\[ v(t) = -A \omega \sin ( \omega t + \phi ) \]
Expert Answer
The given information about this problem is as follows:
\[ \omega = 4.63\ s^{-1} \]
\[ A \omega = 2.60\ cm/s \]
\[ \phi = \pi/2 \]
\[ m = 0.500 kg \]
a) We have the value of $\omega$, so we can use its value to find the time period of the SHM. The time period T is given as:
\[ T = \dfrac{ 2 \pi }{ \omega } \]
Substituting the values, we get:
\[ T = \dfrac{ 2 \pi }{ 4.63 } \]
\[ T = 1.36\ s \]
b) The given equation of the velocity above shows that the constant A before the $\sin$ represents the amplitude. Comparing the equation with the given equation of the velocity of the SHM, we get:
\[ A \omega = 2.60\ cm/s \]
\[ A = \dfrac{ 2.60 \times 10^ {-2} }{ 4.63 s^{-1} } \]
\[ A = 5.6\ mm \]
c) The maximum acceleration of the mass in SHM is given by the equation as:
\[ a_{max} = A \times \omega^2 \]
Substituting the values, we get:
\[ a_{max} = 5.6 \times 10^{-3} \times (4.63)^2 \]
Simplifying the equation, we get:
\[ a_{max} = 0.12 m/s^2 \]
d) The force constant of the spring can be calculated by the given equation as:
\[ \omega = \sqrt{ \dfrac{ k }{ m } } \]
Rearranging the equation to solve for k, we get:
\[ k = m \omega^2 \]
Substituting the values, we get:
\[ k = 0.500 \times (4.63)^2 \]
\[ k = 10.72\ kg/s^2 \]
Numerical Result
a) Time Period:
\[ T = 1.36\ s \]
b) The Amplitude:
\[ A = 5.6\ mm \]
c) Maximum Acceleration:
\[ a_{max} = 0.12 m/s^2 \]
d) Force Constant of the Spring:
\[ k = 10.72\ kg/s^2 \]
Example
A mass is attached to a spring and oscillates, making it a simple harmonic motion. The equation of the velocity is given as follows. Find the amplitude and time period of the SHM.
\[ v_x(t) = ( 4.22 cm/s) \sin \big[ ( 2.74 rad/s ) t – (\pi) \big] \]
The value of the $\omega$ is given as:
\[ \omega = 2.74\ s^{-1} \]
The amplitude A is given as:
\[ A \omega = 4.22 \times 10^{-2} m/s \]
\[ A = \dfrac{ 4.22 \times 10^{-2} }{ 2.74 } \]
\[ A = 15.4\ mm \]
The value of the time period of the SHM is given as:
\[ T = \dfrac{ 2 \pi }{ \omega } \]
\[ T = \dfrac{ 2 \pi }{ 2.74 } \]
\[ T = 2.3\ s \]