(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$. Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant. $k = \frac

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(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$. Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant. $k = \frac{m_1 g}{l_1} = \frac{1.0 \times 10}{0.05} = 200 \,N/m$. Now, for a mass $m_2 = 2.0 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$. $\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 \,rad/s$. The block is pulled by $A = 10 \,cm = 0.1 \,m$, which represents the amplitude of the oscillation. The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$. $v_{\max} = 0.1 \,m \times 10 \,rad/s = 1 \,m/s$.

Class 11 Physics Oscillations SHM of Spring Mass System

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When a body of mass $1.0 \,kg$ is suspended from a certain light spring hanging vertically, its length increases by $5 \,cm$. By suspending a $2.0 \,kg$ block to the spring and if the block is pulled through $10 \,cm$ and released, the maximum velocity in $m/s$ is: (Acceleration due to gravity $= 10 \,m/s^2$)

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SHM of Spring Mass System

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$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)

MediumAP EAMCET 2024

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Two identical springs are connected to a mass $m$ as shown in the figure ($k$ = spring constant). If the time period of the configuration in $(a)$ is $2 \,s$, what is the time period of the configuration in $(b)$?

MediumWBJEE 2011

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$A$ 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 $s$)

MediumNEET 2021

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$A$ mass $m$ is attached to two springs of same force constant $K$,as shown in the following four arrangements. If $T_1, T_2, T_3$ and $T_4$ are the time periods of oscillation in the respective arrangements,in which case is the time period maximum?

Easy

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Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

DifficultAIIMS 2019

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Two identical springs are connected to a mass $m$ as shown in the figure ($k$ = spring constant). If the time period of the configuration in $(a)$ is $2 \,s$, what is the time period of the configuration in $(b)$?

$A$ 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 $s$)

$A$ mass $m$ is attached to two springs of same force constant $K$,as shown in the following four arrangements. If $T_1, T_2, T_3$ and $T_4$ are the time periods of oscillation in the respective arrangements,in which case is the time period maximum?

Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

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