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This is one of our many quizzes on Simple harmonic motion. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Simple harmonic motion Quiz 2

physics

 Total energy of a string of a guitar, which executes simple harmonic motion is in such a way that it gets equally shared to both of its contributing components at a particular point of time. What can be said about the position of this string at this instant, if the crest / trough of the wave motion of this string has a length given by $k$, from the equilibrium position?  

 Recall the expressions for components of total energy of a particle executing SHM and use the given condition to arrive at the answer. 

  $\text{Total energy is given by}\\TE=KE+PE\\\text{At a particular point of consideration, }\\PE=KE \\\Rightarrow \dfrac{1}{2}m{{\omega }^{2}}{{x}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}({{k}^{2}}-{{x}^{2}}) \\\Rightarrow {{x}^{2}}={{k}^{2}}-{{x}^{2}} \\\Rightarrow 2{{x}^{2}}={{k}^{2}} \\\therefore x=\pm \dfrac{k}{\sqrt{2}}$

 The time required for a body executing simple harmonic motion to complete one oscillation is given by $2\pi \sqrt{\dfrac{p}{b}}$. If the amplitude of motion is given by $\sqrt{\dfrac{c}{b}}$, which one of the following is true for two of its velocities corresponding to two different displacements? 

 Recall the expressions for period and amplitude of oscillation of a body executing simple harmonic motion and relate them with the given expressions. 

  Given $T=2\pi \sqrt{\dfrac{p}{b}} $ and $a=\sqrt{\dfrac{c}{b}}$ $ \\ $ We know that time period of oscillation of a body executing simple harmonic motion is given by $ \\T=2\pi \sqrt{\dfrac{({{x}_{2}}^{2}-{{x}_{1}}^{2})}{({{v}_{1}}^{2}-{{v}_{2}}^{2})}}\to (1)\\ $ Amplitude of oscillation is given by $ \\a=\sqrt{\dfrac{({{v}_{1}}^{2}{{x}_{2}}^{2}-{{v}_{2}}^{2}{{x}_{1}}^{2})}{({{v}_{1}}^{2}-{{v}_{2}}^{2})}}\to (2)\\ $ Relating $(1)$ and $(2)$ with the given expressions, we can say that $ \\p={{x}_{2}}^{2}-{{x}_{1}}^{2} \\\Rightarrow b={{v}_{1}}^{2}-{{v}_{2}}^{2} \\\therefore c={{v}_{1}}^{2}{{x}_{2}}^{2}-{{v}_{2}}^{2}{{x}_{1}}^{2}$

 If $\pi \text{ secon}{{\text{d}}^{-1}}$ is the angular frequency of a suspended bead executing simple harmonic motion, what would be its frequency? 

 Determine the period of oscillation of the suspended bead to arrive at the required answer easily.

 We know that angular frequency of an object executing simple harmonic motion is given by $ \\ \omega =\dfrac{2\pi }{T}\\\Rightarrow\omega =\pi \text{ secon}{{\text{d}}^{-1}} \\\Rightarrow \dfrac{2\pi }{T}=\pi \text{ } \\\Rightarrow T=2\text{ seconds} \\ $ Clearly, frequency of oscillation of the suspended bead would be $\\ f=\dfrac{1}{T} \\\therefore f=\dfrac{1}{2}Hz$

 The kinetic and potential energies of an object executing simple harmonic motion are $0.0075\text{ J}$ and $0.0025\text{ J}$, respectively, at a particular instant. If the displacement of this object at this time is equal to $5\text{ cm}$, what would be the amplitude of oscillation? 

 Use the expressions for kinetic and potential energies in simple harmonic motion to derive a ratio and further arrive at the amplitude of oscillation. 

 Given $KE=0.0075\text{ J}$ and $PE=0.0025\text{ J}$ $ \\ $ In simple harmonic motion, we know that $\\KE=\dfrac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\to (1)\\PE=\dfrac{1}{2}m{{\omega }^{2}}{{x}^{2}}\to (2) \\ $ Dividing $(1)$ by $(2)$, we have $ \\ \dfrac{KE}{PE}=\dfrac{{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}} \\\Rightarrow \dfrac{{{a}^{2}}-{{x}^{2}}}{{{x}^{2}}}=\dfrac{0.0075}{0.0025}=3 \\\Rightarrow 3{{x}^{2}}={{a}^{2}}-{{x}^{2}} \\\Rightarrow 4{{x}^{2}}={{a}^{2}} \\\therefore a=2x$

 Fill in the blanks: For a particle executing simple harmonic motion, if the ratio $\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{2}$, ____ represents the time required by the particle to move from the ____ position to the point midway between the stable position and the maximal position whereas ____ represents the time required by the particle to move from the ____ position to the point midway between the stable position and the maximal position. 

 Predict the given periods first, determine them by deriving expressions for the given conditions and finally check them with the given ratio to prove the prediction.  

 Suppose that ${{T}_{1}}$ represents the time required by the particle to move from the stable position to the point midway between the stable position and the maximal position while ${{T}_{2}}$ represents the time required by the particle to move from the maximal position to the point midway between the stable position and the maximal position. $ \\ $ Now, let us determine what ${{T}_{1}}$ and ${{T}_{2}}$ are. $ \\ $ To determine ${{T}_{1}}$ we require: $x=0$ at ${{T}_{1}}=0$ $x=a\sin \omega {{T}_{1}}$  Now, time when $x=\dfrac{a}{2}$ is given by: $\\\dfrac{a}{2}=a\sin \omega {{T}_{1}} \\\Rightarrow \sin \omega {{T}_{1}}=\dfrac{1}{2} \\\Rightarrow \omega {{T}_{1}}=\dfrac{\pi }{6} \\\Rightarrow \left( \dfrac{2\pi }{T} \right){{T}_{1}}=\dfrac{\pi }{6} \\\Rightarrow {{T}_{1}}=\dfrac{T}{12} \\ $  Let this be equation 1. To determine ${{T}_{2}}$ we require: $\\x=a\,\text{at}\,{{T}_{2}}=0\\x=a\cos \omega {{T}_{2}}\\ $ Now, time when $x=\dfrac{a}{2}$ is given by: $\\\dfrac{a}{2}=a\cos \omega {{T}_{2}} \\\Rightarrow \cos \omega {{T}_{2}}=\dfrac{1}{2} \\\Rightarrow \omega {{T}_{2}}=\dfrac{\pi }{3} \\\Rightarrow \left( \dfrac{2\pi }{T} \right){{T}_{2}}=\dfrac{\pi }{3} \\\Rightarrow {{T}_{2}}=\dfrac{T}{6} \\ $ Let this be equation 2. Clearly, from 1 and 2, we have $\\\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{2}\\ $ Therefore, for a particle executing simple harmonic motion, if the ratio $\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{2}$      