$$ \\\\ $$ As the car is moving with a constant velocity, the system of car and pendulum are not accelerating. So as $$a=0$$, there is no pseudo force acting on the bob of the pendulum. $$ \\\\ {{g}_{\\text{effective}}}=\\sqrt{{{g}^{2}}+{{a}^{2}}}\\,\\left( \\therefore \\text{But }a=0\\text{ here as car is moving with constant velocity} \\right) \\\\ \\Rightarrow\\,\\,{{g}_{\\text{effective}}}=g \\\\ $$ Time period, T of the pendulum outside the car, $$ \\\\T=2\\pi\\sqrt{\\dfrac{\\ell }{g}} \\\\ $$ Time period, T of a pendulum in the car, $$ \\\\ \\Rightarrow {{T}_{car}}=2\\pi\\sqrt{\\dfrac{\\ell }{{{g}_{effective}}}} \\\\ \\Rightarrow \\,\\,{{T}_{car}}=2\\pi\\sqrt{\\dfrac{\\ell }{g}} \\\\ \\text{T=}{{\\text{T}}_{car}} \\\\ $$ So, time period is same for both cases.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Period of a pendulum Quiz 1","text":"For a simple pendulum, the time period is $$\\pi $$ seconds. Its velocity is $$10\\text{ m/s}$$ at the instant when it crosses its mean position. The amplitude of oscillation is ","comment":{"@type":"Comment","text":" ${{V}_{\\max }}$ about mean position is $A\\omega $ where $''A''$ is amplitude and $\\omega $ is angular velocity of rotation. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$10\\,m$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$1\\,m$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$20\\,m$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\,m$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$As\\,\\,T=\\pi , \\\\ \\omega =\\dfrac{2\\pi }{T}=\\dfrac{2\\pi }{\\pi }=2\\text{ }rad/\\sec \\\\\\text{For an oscillating particle like a pendulum,} \\\\ \\text{Velocity V is} \\\\ \\text{V(x)=}\\omega \\sqrt{{{A}^{2}}-{{x}^{2}}}, \\\\ \\text{At the mean position x=0,} \\\\ \\Rightarrow \\,\\,\\text{V(0)=}\\omega \\text{A} \\\\ \\Rightarrow \\,\\,\\text{10=2}\\times \\text{A} \\\\ \\therefore \\,\\,\\text{A=5 m}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Period of a pendulum Quiz 1","text":"Two bobs of mass $${{m}_{1}}=10\\,kg$$, $${{m}_{2}}=20\\,kg$$ are oscillated about a fixed point. If the length of the string of bob $$\\left( 1 \\right)$$ is $${{\\ell }_{1}}$$ and that of bob $$\\left( 2 \\right)$$ is $$\\dfrac{{{\\ell }_{1}}}{2}$$ , then the time period of oscillation would be ","comment":{"@type":"Comment","text":"Time period of simple pendulum is $T=2\\pi \\sqrt{\\dfrac{\\ell }{g}}$, where $\\ell $ is length of string and $g$ is effective gravity on the bob. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" more for $${{m}_{2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" equal for both","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None is current","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" more for $${{m}_{1}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" The time period of a simple pendulum of length $$\\ell$$ is $$ \\\\T=2\\pi \\sqrt{\\dfrac{\\ell }{g}} \\\\ $$ Thus, $$ T\\propto \\sqrt{\\ell } $$. As gravitational acceleration, $$g$$ is same for both bobs, we can see that the time period $$T$$ depends only upon the length of the pendulum. $$ \\\\ $$ Further, we can see that the mass of the pendulum does not affect the time period. $$ \\\\ $$ If $${T}_{1}$$ and $${T}_{2}$$ be the time periods and $${{\\ell }_{1}}, {{\\ell }_{2}}$$ be the lengths of bobs $$(1)$$ and $$(2)$$ respectively, then $$ \\\\ \\dfrac{{{T}_{1}}}{{{T}_{2}}}=\\sqrt{\\dfrac{{{\\ell }_{1}}}{{{\\ell }_{2}}}}\\\\ \\Rightarrow\\dfrac{{{T}_{1}}}{{{T}_{2}}} =\\sqrt{\\dfrac{{{\\ell }_{1}}}{\\dfrac{{{\\ell }_{1}}}{2}}}\\\\ \\Rightarrow \\dfrac{{{T}_{1}}}{{{T}_{2}}}=\\sqrt{2} \\\\ \\Rightarrow \\,\\,{{T}_{1}}=\\sqrt{2}{{T}_{2}} \\\\ \\therefore \\,\\,{{T}_{1}} > {{T}_{2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Period of a pendulum Quiz 1","text":"A simple pendulum oscillates about a common point. The pendulum has a mass of $$1\\,kg$$ and an oscillation duration of $$2\\pi $$ seconds. Velocity at the lowest point is $$100\\,m/s$$. Then the maximum potential energy that the pendulum can store is. ","comment":{"@type":"Comment","text":"${{V}_{\\max }}=a\\omega $ at mean position of oscillation. The $K.E$ is maximum and potential energy is zero at this instant, while the total Mechanical energy is conserved. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{10}^{4}}\\,Joule$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\,\\times {{10}^{4}}\\,Joule$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $${{10}^{5}}\\,Joule$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\times {{10}^{3}}\\,Joule$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ \\text{For oscillation of a simple pendulum,}\\\\ \\text{the total mechanical energy is conserved}\\text{.} \\\\K.E\\text{ at mean position}=\\dfrac{1}{2}m{{v}^{2}}=\\dfrac{1}{2}\\times 1\\times 100\\times 100 =5000\\text{ Joule} \\\\ \\text{If we consider the gravitational reference potential line to be at the lowest i.e,}\\\\ \\text{Mean position of the pendulum, } \\\\ \\text{then }P.E\\text{ at mean position}=Mgh=1\\times 10\\times 0=0\\,\\,Joule. \\\\ \\text{In a pendulum, the point where }K.E\\text{ is zero, }\\\\ P.E\\text{ becomes maximum (as total Mechanical energy is conserved)}\\text{.} \\\\ \\text{Thus,} K.{{E}_{mean }}+P.{{E}_{mean }}=K.{{E}_{\\max }}+P.{{E}_{\\max }} \\\\ \\Rightarrow 5000+0=0+P.{{E}_{\\max }} \\\\ \\therefore P.{{E}_{\\max }}=5000\\text{ Joule}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Period of a pendulum Quiz 1","text":"If the length of a pendulum doubles then its time period changes by $$20$$ seconds. The initial time period of pendulum is","comment":{"@type":"Comment","text":"Time period of the pendulum is related with length and gravity as $T=2\\pi \\sqrt{\\dfrac{\\ell }{g}}$.Use it to solve the problem. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$20\\sec $$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$60\\sec $$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$90\\sec $$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$40\\sec $$","position":1,"answerExplanation":{"@type":"Comment","text":" $$For\\text{ a pendulum, }T=2\\pi \\sqrt{\\dfrac{\\ell }{g}} \\\\As\\text{ gravitational acceleration g is same in both cases,} \\\\ \\Rightarrow T\\propto \\sqrt{\\ell } \\\\ \\Rightarrow \\,\\,\\dfrac{\\Delta T}{T}\\times 100=\\dfrac{1}{2}\\dfrac{\\Delta L}{L}\\times 100 \\\\ \\Rightarrow\\dfrac{\\Delta T}{T}\\times 100=\\dfrac{1}{2}\\times \\dfrac{2L-L}{L}\\times 100 \\\\ \\Rightarrow \\dfrac{\\Delta T}{T}=0.5 \\\\ \\Rightarrow \\dfrac{{{T}_{f}}-{{T}_{i}}}{{{T}_{i}}}=0.5 \\\\ \\therefore {{T}_{i}}=40\\,\\,\\text{sec}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}