Question:
Position of a \(3\ \text{kg}\) mass moving along the \(x\)-axis is given by \(x=0.3\cos(\omega t)\ \text{m}\). If \(K(t)\) denotes the kinetic energy at time \(t\), then the value of \(\dfrac{K\left(\frac{\pi}{6\omega}\right)}{K\left(\frac{\pi}{3\omega}\right)}\) is
Position of a \(3\ \text{kg}\) mass moving along the \(x\)-axis is given by \(x=0.3\cos(\omega t)\ \text{m}\). If \(K(t)\) denotes the kinetic energy at time \(t\), then the value of \(\dfrac{K\left(\frac{\pi}{6\omega}\right)}{K\left(\frac{\pi}{3\omega}\right)}\) is
Show Hint
For SHM given by
\[
x=A\cos(\omega t),
\]
velocity is
\[
v=-A\omega\sin(\omega t),
\]
so kinetic energy varies as
\[
K\propto \sin^2(\omega t).
\]
Updated On: Jun 26, 2026
- \(\dfrac{1}{3}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\sqrt{3}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the given displacement equation.
The position of the mass is \[ x=0.3\cos(\omega t) \] This is of the form \[ x=A\cos(\omega t) \] where \[ A=0.3\ \text{m} \]
Step 2: Find the velocity.
Velocity is given by \[ v=\frac{dx}{dt} \] So, \[ v=\frac{d}{dt}\left(0.3\cos(\omega t)\right) \] \[ v=-0.3\omega\sin(\omega t) \]
Step 3: Write the expression for kinetic energy.
Kinetic energy is \[ K(t)=\frac{1}{2}mv^2 \] Substituting the value of \(v\), \[ K(t)=\frac{1}{2}m(0.3\omega)^2\sin^2(\omega t) \] Thus, \[ K(t)\propto \sin^2(\omega t) \]
Step 4: Calculate \(K\left(\frac{\pi}{6\omega}\right)\).
For \[ t=\frac{\pi}{6\omega} \] we get \[ \omega t=\omega \cdot \frac{\pi}{6\omega} \] \[ \omega t=\frac{\pi}{6} \] Therefore, \[ K\left(\frac{\pi}{6\omega}\right)\propto \sin^2\frac{\pi}{6} \] Since \[ \sin\frac{\pi}{6}=\frac{1}{2} \] we get \[ K\left(\frac{\pi}{6\omega}\right)\propto \left(\frac{1}{2}\right)^2 \] \[ K\left(\frac{\pi}{6\omega}\right)\propto \frac{1}{4} \]
Step 5: Calculate \(K\left(\frac{\pi}{3\omega}\right)\).
For \[ t=\frac{\pi}{3\omega} \] we get \[ \omega t=\omega \cdot \frac{\pi}{3\omega} \] \[ \omega t=\frac{\pi}{3} \] Therefore, \[ K\left(\frac{\pi}{3\omega}\right)\propto \sin^2\frac{\pi}{3} \] Since \[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \] we get \[ K\left(\frac{\pi}{3\omega}\right)\propto \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ K\left(\frac{\pi}{3\omega}\right)\propto \frac{3}{4} \]
Step 6: Find the required ratio.
Now, \[ \frac{K\left(\frac{\pi}{6\omega}\right)} {K\left(\frac{\pi}{3\omega}\right)} = \frac{\frac{1}{4}}{\frac{3}{4}} \] \[ = \frac{1}{4}\times \frac{4}{3} \] \[ =\frac{1}{3} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{1}{3}} \]
The position of the mass is \[ x=0.3\cos(\omega t) \] This is of the form \[ x=A\cos(\omega t) \] where \[ A=0.3\ \text{m} \]
Step 2: Find the velocity.
Velocity is given by \[ v=\frac{dx}{dt} \] So, \[ v=\frac{d}{dt}\left(0.3\cos(\omega t)\right) \] \[ v=-0.3\omega\sin(\omega t) \]
Step 3: Write the expression for kinetic energy.
Kinetic energy is \[ K(t)=\frac{1}{2}mv^2 \] Substituting the value of \(v\), \[ K(t)=\frac{1}{2}m(0.3\omega)^2\sin^2(\omega t) \] Thus, \[ K(t)\propto \sin^2(\omega t) \]
Step 4: Calculate \(K\left(\frac{\pi}{6\omega}\right)\).
For \[ t=\frac{\pi}{6\omega} \] we get \[ \omega t=\omega \cdot \frac{\pi}{6\omega} \] \[ \omega t=\frac{\pi}{6} \] Therefore, \[ K\left(\frac{\pi}{6\omega}\right)\propto \sin^2\frac{\pi}{6} \] Since \[ \sin\frac{\pi}{6}=\frac{1}{2} \] we get \[ K\left(\frac{\pi}{6\omega}\right)\propto \left(\frac{1}{2}\right)^2 \] \[ K\left(\frac{\pi}{6\omega}\right)\propto \frac{1}{4} \]
Step 5: Calculate \(K\left(\frac{\pi}{3\omega}\right)\).
For \[ t=\frac{\pi}{3\omega} \] we get \[ \omega t=\omega \cdot \frac{\pi}{3\omega} \] \[ \omega t=\frac{\pi}{3} \] Therefore, \[ K\left(\frac{\pi}{3\omega}\right)\propto \sin^2\frac{\pi}{3} \] Since \[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \] we get \[ K\left(\frac{\pi}{3\omega}\right)\propto \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ K\left(\frac{\pi}{3\omega}\right)\propto \frac{3}{4} \]
Step 6: Find the required ratio.
Now, \[ \frac{K\left(\frac{\pi}{6\omega}\right)} {K\left(\frac{\pi}{3\omega}\right)} = \frac{\frac{1}{4}}{\frac{3}{4}} \] \[ = \frac{1}{4}\times \frac{4}{3} \] \[ =\frac{1}{3} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{1}{3}} \]
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