Oscillations

Explanation

The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:

$n=\frac{12.5}{2}=6.25\text{ cycles. }$

In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:

$D=6.25\times 4A=6.25\times 4\times 1=25cm$

Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.

$\frac{D}{d}=\frac{25}{1}=25$

Explanation

$g_1=\frac{GM}{{\left(R+R\right)}^2}$

$g_2=\frac{GM}{{\left(R+2R\right)}^2}$

$\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}=\frac{2}{3}$

$\Rightarrow 3T_1=2T_2$

ExplanationStep 1: Understanding SHM and Maximum Speed

For a simple harmonic motion (SHM), the maximum speed $V_{max}$​ of a particle is given by:

$V_{max}=A\omega$

where:

• $A$ is the amplitude of oscillation.

• $\omega$ is the angular frequency, given by:

  $$\omega = \sqrt{\frac{k}{m}}$$​

Thus, the maximum speed becomes:

$V_{max}=A\sqrt{\frac{k}{m}}$​

Step 2: Finding the Ratio of Maximum Speeds

Let the two particles have spring constants $K_1$​ and $K_2$ but the same mass $m$ and same amplitude $A$.

For the first particle:

$$v_{\max1} = A \sqrt{\frac{K_1}{m}}$$​

For the second particle:

$$v_{\max2} = A \sqrt{\frac{K_2}{m}}$$

Taking the ratio:

$$\frac{v_{\max1}}{v_{\max2}} = \frac{A \sqrt{\frac{K_1}{m}}}{A \sqrt{\frac{K_2}{m}}}$$​​

Canceling $A$A and simplifying:

$$\frac{v_{\max1}}{v_{\max2}} = \sqrt{\frac{K_1}{K_2}}$$

Step 3: Selecting the Correct Option

$$\boxed{A} \quad \sqrt{\frac{K_1}{K_2}}$$

Explanation

$$\begin{gathered} T=2\pi \sqrt{\frac{\ell }{g}} \\ g=\frac{g_0{R}^2}{(R+h{)}^2} \end{gathered}$$

As h increases, g decreases, and T increases.

Explanation

\begin{equation}
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\
& \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \\
& \text { Percentage error, } \frac{\Delta \mathrm{g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \\
& =\frac{0.2}{20}+2\left(\frac{1}{40}\right)=\frac{1.2}{20}
\end{aligned}
\end{equation}

percentage change = $\frac{1.2}{20}\times 100=6\%$

Explanation

1. Key concepts:

   • The total distance $D$ is the sum of all oscillations' displacements (absolute values of motion).
   • Displacement $d$d is the shortest distance between the initial and final positions.
   • The time period $T = 2 \, \text{s}$.

2. Oscillation in 12.5 seconds:

   • Number of complete oscillations: $$n = \frac{\text{Total time}}{\text{Time period}} = \frac{12.5}{2} = 6.25$$
   • This implies 6 full oscillations and 1/4 of another oscillation.

3. Displacement ($d$):

   • After 6 full oscillations, the block returns to its starting position.
   • During the remaining $0.25 \times T = 0.5 \, \text{s}$, the block moves from the mean position to the maximum displacement (amplitude $A$).
   • Hence, $d = A$

4. Total distance ($D$):

   • In one complete oscillation, the block covers a distance of $4A$(2A forward and 2A backward).
   • For 6 complete oscillations: $$D_{\text{6 oscillations}} = 6 \cdot 4A = 24A$$
   • In the last quarter oscillation: $$D_{\text{last quarter}} = A$$
   • Total distance: $$D = D_{\text{6 oscillations}} + D_{\text{last quarter}} = 24A + A = 25A$$

5. Ratio $\frac{D}{d}$​:

   $$\frac{D}{d} = \frac{25A}{A} = 25$$

Answer: $\frac{D}{d} = 25$

Explanation

y = 2x. This is the equation of a straight line. Since x and y depend on cos(ωt), the motion is oscillatory (back and forth) along this line segment between (–A, –2A) and (A, 2A). This is Simple Harmonic Motion (SHM) and also Periodic Motion.

(NEW NCERT 11^th Page No. 260)

Explanation

1. Key concepts:

   • The total distance $D$ is the sum of all oscillations' displacements (absolute values of motion).
   • Displacement $d$d is the shortest distance between the initial and final positions.
   • The time period $T = 2 \, \text{s}$.

2. Oscillation in 12.5 seconds:

   • Number of complete oscillations: $$n = \frac{\text{Total time}}{\text{Time period}} = \frac{12.5}{2} = 6.25$$
   • This implies 6 full oscillations and 1/4 of another oscillation.

3. Displacement ($d$):

   • After 6 full oscillations, the block returns to its starting position.
   • During the remaining $0.25 \times T = 0.5 \, \text{s}$, the block moves from the mean position to the maximum displacement (amplitude $A$).
   • Hence, $d = A$

4. Total distance ($D$):

   • In one complete oscillation, the block covers a distance of $4A$(2A forward and 2A backward).
   • For 6 complete oscillations: $$D_{\text{6 oscillations}} = 6 \cdot 4A = 24A$$
   • In the last quarter oscillation: $$D_{\text{last quarter}} = A$$
   • Total distance: $$D = D_{\text{6 oscillations}} + D_{\text{last quarter}} = 24A + A = 25A$$

5. Ratio $\frac{D}{d}$​:

   $$\frac{D}{d} = \frac{25A}{A} = 25$$

Answer: $\frac{D}{d} = 25$