The maximum velocity of a particle, executing simple harmonic motion with an amplitude \[7\;mm\], is \[4.4\;m/s\]. The period of oscillation is:
A. \[100s\]
B. \[0.01s\]
C. \[10s\]
D. \[0.1s\]
Answer
504k+ views
Hint: Simple harmonic motion is a form of a motion in which the displacement of the body from its mean position is directly proportional restoring force. To answer this question we need to use the formula for displacement in simple harmonic motion as a function of time. In the formula, we need to substitute the given values and we will get the answer that we need.
Complete step by step solution:
In simple harmonic motion the displacement as a function of time we get,
\[\;x = Asin(\omega t + \phi )\]
Here \[A\]is the amplitude of the
\[\omega \] is the angular frequency
\[t\] is the time taken for the object in motion to complete one oscillation
Now to get the velocity we need to differentiate the displacement equation.
Therefore differentiating the above equation we get,
Velocity \[v = \dfrac{{dx}}{{dt}} = A\omega cos(\omega t + \phi )\]
Here velocity will be maximum when \[{v_{\max }} = A\omega \]
We know that the angular velocity \[\omega = \dfrac{{2\pi }}{T}\]
Given the maximum velocity of the particle, \[{v_{\max }} = 4.4\;m/s\]
Amplitude \[A = 7\;mm = 7 \times {10^{ - 3}}m\]
Substituting all the values in \[{v_{\max }} = A\omega \]
we get,
\[4.4\; = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{T}\]
\[ \Rightarrow T = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{{4.4\;}}\]
\[ \Rightarrow T = 9.99 \times {10^{ - 3}}\]
\[ \Rightarrow T \simeq 10 \times {10^{ - 3}}\]
\[ \therefore T \simeq 0.01s\]
Therefore the correct option is B.
Note:
The angular frequency ω is given by \[\omega = \dfrac{{2\pi }}{T}\]. The angular frequency is measured in terms of radians per second. If we take the inverse of the period then we will get its frequency \[f = \dfrac{1}{T}\] . The frequency \[f = \dfrac{1}{T} = \dfrac{\omega }{{2\pi }}\] of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz, \[\left( {1Hz = \dfrac{1}{s}} \right)\].The quantity \[\phi \] is said to be the phase constant. It can be determined using the initial conditions of the motion. If we consider at \[t = 0\] the object will have its maximum displacement along the positive x-direction, then \[\phi = 0\], if the object has its maximum displacement along the negative x-direction, then \[\phi = \pi \]. If the particle is moving through its equilibrium position with the maximum velocity at \[t = 0\] in the negative x-direction then \[\phi = \dfrac{\pi }{2}\]. Then the quantity \[\left( {\omega t + \phi {\text{ }}} \right)\] is said to be the phase.
Complete step by step solution:
In simple harmonic motion the displacement as a function of time we get,
\[\;x = Asin(\omega t + \phi )\]
Here \[A\]is the amplitude of the
\[\omega \] is the angular frequency
\[t\] is the time taken for the object in motion to complete one oscillation
Now to get the velocity we need to differentiate the displacement equation.
Therefore differentiating the above equation we get,
Velocity \[v = \dfrac{{dx}}{{dt}} = A\omega cos(\omega t + \phi )\]
Here velocity will be maximum when \[{v_{\max }} = A\omega \]
We know that the angular velocity \[\omega = \dfrac{{2\pi }}{T}\]
Given the maximum velocity of the particle, \[{v_{\max }} = 4.4\;m/s\]
Amplitude \[A = 7\;mm = 7 \times {10^{ - 3}}m\]
Substituting all the values in \[{v_{\max }} = A\omega \]
we get,
\[4.4\; = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{T}\]
\[ \Rightarrow T = 7 \times {10^{ - 3}} \times \dfrac{{2\pi }}{{4.4\;}}\]
\[ \Rightarrow T = 9.99 \times {10^{ - 3}}\]
\[ \Rightarrow T \simeq 10 \times {10^{ - 3}}\]
\[ \therefore T \simeq 0.01s\]
Therefore the correct option is B.
Note:
The angular frequency ω is given by \[\omega = \dfrac{{2\pi }}{T}\]. The angular frequency is measured in terms of radians per second. If we take the inverse of the period then we will get its frequency \[f = \dfrac{1}{T}\] . The frequency \[f = \dfrac{1}{T} = \dfrac{\omega }{{2\pi }}\] of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz, \[\left( {1Hz = \dfrac{1}{s}} \right)\].The quantity \[\phi \] is said to be the phase constant. It can be determined using the initial conditions of the motion. If we consider at \[t = 0\] the object will have its maximum displacement along the positive x-direction, then \[\phi = 0\], if the object has its maximum displacement along the negative x-direction, then \[\phi = \pi \]. If the particle is moving through its equilibrium position with the maximum velocity at \[t = 0\] in the negative x-direction then \[\phi = \dfrac{\pi }{2}\]. Then the quantity \[\left( {\omega t + \phi {\text{ }}} \right)\] is said to be the phase.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

