Unit 7: Oscillations
Weightage : 10-15%
The figure above shows a particle executing uniform circular motion in a circle of radius \(R\). Light sources (not shown) cause shadows of the particle to be projected onto two mutually perpendicular screens. The positive directions for \(x\) and \(y\) along the screens are denoted by the arrows.
When the shadow on Screen 1 is at position \(x=-0.5R\) and moving in the \(+x\) direction, what is true about the position and velocity of the shadow on Screen 2 at that same instant?
Correct Answer: \( \boxed{\mathrm{A}} \)
The shadow coordinates are the projections of the particle’s position onto the coordinate axes:
\(x=R\cos\theta,\qquad y=R\sin\theta\)
Given
\(x=-0.5R\),
we have
\(\cos\theta=-0.5\).
Since the shadow on Screen 1 is moving in the \(+x\) direction, the particle is in the third quadrant, where
\(\theta=240^\circ\).
Therefore,
\(y=R\sin240^\circ=-\dfrac{\sqrt{3}}{2}R\approx-0.866R\).
As the particle continues its counterclockwise motion, the \(y\)-coordinate becomes more negative, so the shadow on Screen 2 moves in the \(-y\) direction.
Hence, the correct answer is (A).
An unstretched ideal spring hangs vertically from a fixed support. A \(0.4\,\mathrm{kg}\) object is then attached to the lower end of the spring. The object is pulled down to a distance of \(0.35\,\mathrm{m}\) below the unstretched position and released from rest at time \(t=0\). A graph of the subsequent vertical position \(y\) of the lower end of the spring as a function of \(t\) is shown above, where \(y=0\) when the spring was initially unstretched.
At which of the following times is the upward velocity of the object the greatest?
Correct Answer: \( \boxed{\mathrm{B}} \)
In simple harmonic motion, the speed is greatest as the object passes through its equilibrium position and zero at the turning points.
The object is released from rest at the lowest point at \(t=0\), so its initial velocity is zero.
From the graph, the object first passes through the equilibrium position while moving upward at approximately \(t=0.25\,\mathrm{s}\). At this instant, the magnitude of the velocity is maximum and its direction is upward.
At \(t=0.50\,\mathrm{s}\), the object is at the upper turning point, so the velocity is again zero. At \(t=0.75\,\mathrm{s}\), it is moving downward.
Therefore, the upward velocity is greatest at \(t=0.25\,\mathrm{s}\), so the correct answer is (B).
A spring-mass system is vibrating along a frictionless, horizontal floor. The spring constant is \(8~\mathrm{N\,m^{-1}}\), the amplitude is \(5~\mathrm{cm}\), and the period is \(4~\mathrm{s}\).
Which of the following equations could represent the position \(x\) of the mass from equilibrium as a function of time \(t\), where \(x\) is in meters and \(t\) is in seconds?
Correct Answer: \( \boxed{\mathrm{C}} \)
The general equation for simple harmonic motion is
\(x=A\cos(\omega t+\phi)\)
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. Since no phase constant is specified, we compare only the amplitude and angular frequency.
The amplitude is
\(A=5~\mathrm{cm}=0.05~\mathrm{m}\)
The angular frequency is related to the period by
\(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}~\mathrm{rad\,s^{-1}}\)
Therefore, the equation of motion is
\(x=0.05\cos\!\left(\dfrac{\pi}{2}t\right)\)
Hence, the correct answer is (C).
