Assuming all pulleys,springs,and strings are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the tension in the string be $T$. From the force balance on the block,the tension $T = mg$.For the pulleys and springs,let the extensions in t

Vedclass · Accepted Answer

The angular frequency for small oscillation of the system is $\sqrt{\frac{4K}{33m}}$.

Vedclass · Answer

(B) Let the tension in the string be $T$. From the force balance on the block,the tension $T = mg$.For the pulleys and springs,let the extensions in the three springs be $x_1, x_2, x_3$ respectively. The effective spring constant $K_{eff}$ is determined by the relation $x = F/K_{eff}$,where $x$ is the displacement of the block.Using the constraint relation for the pulleys,the displacement of the block $x$ is related to the extensions as $x = \frac{x_1}{2} + 2x_2 + 2x_3$.Given the tension $T$ in the string,the forces on the springs are $F_1 = T/2$,$F_2 = 2T$,and $F_3 = 2T$.Thus,$x_1 = \frac{T}{2K}$,$x_2 = \frac{2T}{K}$,and $x_3 = \frac{2T}{K}$.Substituting these into the constraint equation: $x = \frac{T}{4K} + \frac{4T}{K} + \frac{4T}{K} = \frac{T + 16T + 16T}{4K} = \frac{33T}{4K}$.Since $T = mg$,$x = \frac{33mg}{4K}$. The effective spring constant is $K_{eff} = \frac{mg}{x} = \frac{4K}{33}$.The angular frequency $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{4K}{33m}}$.The elastic potential energy at equilibrium is $U = \frac{1}{2} K_1 x_1^2 + \frac{1}{2} K_2 x_2^2 + \frac{1}{2} K_3 x_3^2 = \frac{1}{2} K (\frac{mg}{2K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 = \frac{m^2g^2}{8K} + \frac{2m^2g^2}{K} + \frac{2m^2g^2}{K} = \frac{m^2g^2 + 16m^2g^2 + 16m^2g^2}{8K} = \frac{33m^2g^2}{8K}$.

Assuming all pulleys,springs,and strings are massless. Consider all surfaces smooth. Choose the correct statement$(s)$.

Similar Questions

Two identical springs of spring constant $k$ are attached to a block of mass $m$ and to fixed supports as shown in the figure. Show that when the mass is displaced from its equilibrium position on either side,it executes simple harmonic motion. Find the period of oscillations.

$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)

$A$ block of mass $m$ attached to one end of a vertical spring produces an extension $x$. If the block is pulled and released,the periodic time of oscillation is:

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle,the new time period will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module