CBSE Class 11 Physics Exemplar Chapter 13 Oscillations NCERT Solutions

Solution: The given equation for the displacement of the particle is $y\; =\; 3\backslash cos\backslash left(\backslash frac\{\backslash pi\}\{4\}\; -\; 2\backslash omega\; t\backslash right)$.

We can rewrite this equation by factoring out a negative sign from the argument of the cosine function: $y\; =\; 3\backslash cos\backslash left(-\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\backslash right)$. Since $\backslash cos(-\backslash theta)\; =\; \backslash cos(\backslash theta)$, this becomes $y\; =\; 3\backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)$.

The standard form of a simple harmonic motion (SHM) equation is $y\; =\; A\backslash cos(\backslash omega\text{'}\; t\; +\; \backslash phi)$, where $A$ is the amplitude, $\backslash omega\text{'}$ is the angular frequency, and $\backslash phi$ is the phase constant.

Comparing the given equation with the standard form, we have:

• Amplitude $A\; =\; 3$
• Angular frequency $\backslash omega\text{'}\; =\; 2\backslash omega$
• Phase constant $\backslash phi\; =\; -\backslash frac\{\backslash pi\}\{4\}$

The time period ($T\text{'}$) of SHM is related to the angular frequency by $T\text{'}\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\text{'}\}$. Substituting $\backslash omega\text{'}\; =\; 2\backslash omega$, we get $T\text{'}\; =\; \backslash frac\{2\backslash pi\}\{2\backslash omega\}\; =\; \backslash frac\{\backslash pi\}\{\backslash omega\}$.

To confirm it's SHM, we can check the acceleration. The velocity is $v\; =\; \backslash frac\{dy\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; \backslash left[\; 3\backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\; \backslash right]\; =\; 3\; \backslash left(-2\backslash omega\backslash right)\; \backslash sin\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\; =\; -6\backslash omega\backslash sin\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)$.

The acceleration is $a\; =\; \backslash frac\{dv\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; \backslash left[\; -6\backslash omega\backslash sin\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\; \backslash right]\; =\; -6\backslash omega\; \backslash left(2\backslash omega\backslash right)\; \backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\; =\; -12\backslash omega^2\backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)$.

Since $y\; =\; 3\backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)$, we can write $\backslash cos\backslash left(2\backslash omega\; t\; -\; \backslash frac\{\backslash pi\}\{4\}\backslash right)\; =\; \backslash frac\{y\}\{3\}$. Substituting this into the acceleration equation gives $a\; =\; -12\backslash omega^2\; \backslash left(\backslash frac\{y\}\{3\}\backslash right)\; =\; -4\backslash omega^2\; y$.

Because the acceleration $a$ is proportional to the negative of the displacement $y$ ($a\; \backslash propto\; -y$), the motion is indeed simple harmonic.

Answer: (b) simple harmonic with period $\backslash pi/\backslash omega$

Solution: The given equation for the displacement is $y\; =\; \backslash sin^3\; \backslash omega\; t$.

To analyze the motion, we use the trigonometric identity $\backslash sin\; 3\backslash theta\; =\; 3\backslash sin\; \backslash theta\; -\; 4\backslash sin^3\; \backslash theta$. Rearranging this identity to solve for $\backslash sin^3\; \backslash theta$, we get $\backslash sin^3\; \backslash theta\; =\; \backslash frac\{3\backslash sin\; \backslash theta\; -\; \backslash sin\; 3\backslash theta\}\{4\}$.

Applying this to our displacement equation, we have:

$y\; =\; \backslash frac\{3\backslash sin\; \backslash omega\; t\; -\; \backslash sin\; (3\backslash omega\; t)\}\{4\}$

Now, let's find the acceleration by differentiating twice with respect to time ($t$).

First derivative (velocity, $v\; =\; \backslash frac\{dy\}\{dt\}$):

$v\; =\; \backslash frac\{d\}\{dt\}\; \backslash left[\; \backslash frac\{3\backslash sin\; \backslash omega\; t\; -\; \backslash sin\; (3\backslash omega\; t)\}\{4\}\; \backslash right]\; =\; \backslash frac\{1\}\{4\}\; \backslash left[\; 3(\backslash omega\backslash cos\; \backslash omega\; t)\; -\; (3\backslash omega\backslash cos\; 3\backslash omega\; t)\; \backslash right]$

$v\; =\; \backslash frac\{3\backslash omega\}\{4\}\; (\backslash cos\; \backslash omega\; t\; -\; \backslash cos\; 3\backslash omega\; t)$

Second derivative (acceleration, $a\; =\; \backslash frac\{dv\}\{dt\}$):

$a\; =\; \backslash frac\{d\}\{dt\}\; \backslash left[\; \backslash frac\{3\backslash omega\}\{4\}\; (\backslash cos\; \backslash omega\; t\; -\; \backslash cos\; 3\backslash omega\; t)\; \backslash right]\; =\; \backslash frac\{3\backslash omega\}\{4\}\; \backslash left[\; (-\backslash omega\backslash sin\; \backslash omega\; t)\; -\; (-3\backslash omega\backslash sin\; 3\backslash omega\; t)\; \backslash right]$

$a\; =\; \backslash frac\{3\backslash omega^2\}\{4\}\; (3\backslash sin\; 3\backslash omega\; t\; -\; \backslash sin\; \backslash omega\; t)$

For simple harmonic motion, the acceleration must be directly proportional to the negative of the displacement, i.e., $a\; \backslash propto\; -y$. In our case, $a\; =\; \backslash frac\{3\backslash omega^2\}\{4\}\; (3\backslash sin\; 3\backslash omega\; t\; -\; \backslash sin\; \backslash omega\; t)$ and $y\; =\; \backslash frac\{3\backslash sin\; \backslash omega\; t\; -\; \backslash sin\; 3\backslash omega\; t\}\{4\}$. We can see that $a$ is not directly proportional to $-y$ because of the different arguments ($\backslash omega\; t$ and $3\backslash omega\; t$) and coefficients involved.

Therefore, the motion is not simple harmonic.

However, the expression for $y$ involves sine functions of $\backslash omega\; t$ and $3\backslash omega\; t$. Since these functions repeat at regular intervals, the overall motion is periodic.

Answer: (b) periodic but not simple harmonic

Solution: For a particle to exhibit simple harmonic motion (SHM), its acceleration ($a$) must be directly proportional to the negative of its displacement ($x$) from the equilibrium position. Mathematically, this condition is expressed as $a\; \backslash propto\; -x$, or $a\; =\; -k\; x$, where $k$ is a positive constant.

Let's examine each option:

• Option (a): $a\_x\; =\; +2x$. Here, acceleration is proportional to positive displacement ($a\; \backslash propto\; +x$). This does not represent SHM.
• Option (b): $a\_x\; =\; +2x^2$. Here, acceleration is proportional to the square of positive displacement ($a\; \backslash propto\; +x^2$). This is not SHM.
• Option (c): $a\_x\; =\; -2x^2$. Here, acceleration is proportional to the square of negative displacement ($a\; \backslash propto\; -x^2$). For SHM, the proportionality must be linear with displacement, not its square.
• Option (d): $a\_x\; =\; -2x$. Here, acceleration is proportional to the negative of displacement ($a\; \backslash propto\; -x$), with $k\; =\; 2$. This fits the condition for SHM.

Therefore, the particle described by option (d) is exhibiting simple harmonic motion.

Answer: (d) $a\_x\; =\; -2x$

Solution: Consider a U-tube containing a liquid. When the liquid column is displaced from its equilibrium position by a height $x$ on one side (meaning it is depressed by $x$ on one side and raised by $x$ on the other side, so the total difference in levels is $2x$), a restoring force acts on it due to gravity.

Let the cross-sectional area of the U-tube be $A$ and the density of the liquid be $\backslash rho$. The excess mass on the lower side is $m\; =\; A\; x\; \backslash rho$. The restoring force ($F$) acting downwards is equal to the weight of this excess mass:

$F\; =\; -\; (A\; x\; \backslash rho)\; g$

The negative sign indicates that the force is always directed opposite to the displacement $x$ (which is taken as positive upwards from equilibrium).

According to Newton's second law, $F\; =\; m\_\{total\}\; a$, where $m\_\{total\}$ is the total mass of the oscillating liquid column. Let the total length of the liquid column be $L$. Then $m\_\{total\}\; =\; A\; L\; \backslash rho$.

So, $A\; L\; \backslash rho\; a\; =\; -\; A\; x\; \backslash rho\; g$.

Simplifying this, we get the acceleration $a$:

$a\; =\; -\; \backslash frac\{A\; x\; \backslash rho\; g\}\{A\; L\; \backslash rho\}\; =\; -\; \backslash frac\{g\}\{L\}\; x$

This equation is in the form $a\; =\; -k\; x$, where $k\; =\; \backslash frac\{g\}\{L\}$ is a positive constant. This shows that the acceleration is directly proportional to the negative of the displacement, which is the condition for simple harmonic motion.

The angular frequency ($\backslash omega$) for this SHM is given by $\backslash omega\; =\; \backslash sqrt\{\backslash frac\{k\}\{m\_\{total\}\}\}\; =\; \backslash sqrt\{\backslash frac\{Ag\backslash rho\}\{AL\backslash rho\}\}\; =\; \backslash sqrt\{\backslash frac\{g\}\{L\}\}$. (Note: Alternatively, using $a\; =\; -\backslash omega^2\; x$, we directly get $\backslash omega^2\; =\; \backslash frac\{g\}\{L\}$).

The time period ($T$) is related to the angular frequency by $T\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\}$.

$T\; =\; \backslash frac\{2\backslash pi\}\{\backslash sqrt\{g/L\}\}\; =\; 2\backslash pi\; \backslash sqrt\{\backslash frac\{L\}\{g\}\}$

From the expression for the time period $T\; =\; 2\backslash pi\; \backslash sqrt\{\backslash frac\{L\}\{g\}\}$, we can see that the time period depends on the total length of the liquid column ($L$) and the acceleration due to gravity ($g$), but it does not depend on the density ($\backslash rho$) or the cross-sectional area ($A$) of the liquid.

Therefore, the motion is simple harmonic, and its time period is independent of the density of the liquid.

Answer: (c) simple harmonic and time period is independent of the density of the liquid