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Model Paper 10 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 105 Marks
Question
A spring having with a spring constant $1200 N m ^{-1}$ is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Image

Take the position of mass when the spring is unstreched as $x=0$, and the direction from left to right as the positive direction of $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch $(t=0)$, the mass is
a. at the mean position,
b. at the maximum stretched position, and
c. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
✓

Answer

The functions have the same frequency and amplitude, but different initial phases.
Given:
Distance travelled by the mass sideways is given by, $A =2.0 cm$
Force constant of the spring is given by, $k=1200 N m ^{-1}$
Mass, m is given by $=3 kg$
Angular frequency of oscillation is given by:
$
\begin{aligned}
& \omega=\sqrt{\frac{\text { spring cons } \tan t}{\text { mass }}} \\
& \Rightarrow \omega=\sqrt{\frac{k}{m}} \\
& =\sqrt{\frac{1200}{3}}=\sqrt{400}=20 rads^{-1}
\end{aligned}
$
a. When the mass is at the mean position, the initial phase is 0.
Displacement is given by,
$
\Rightarrow x=A \sin \omega t=2 \sin 20 t
$
b. At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is $\frac{\pi}{2}$. hence, Displacement is given by,
$
\begin{aligned}
& \Rightarrow x=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
& =2 \sin \left(20 t+\frac{\pi}{2}\right)=2 \cos 20 t
\end{aligned}
$
c. At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is $\frac{3 \pi}{2}$.
Displacement is given by,
$
\begin{aligned}
& \Rightarrow x=A \sin \left(\omega t+\frac{3 \pi}{2}\right) \\
& =2 \sin \left(20 t+\frac{3 \pi}{2}\right)=-2 \cos 20 t
\end{aligned}
$
d. The functions have the same frequency $\left(\frac{20}{2 \pi} H z\right)$ and amplitude $(2 cm)$, but initial phases are different $\left(0, \frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

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