What is meant by simple harmonic motion at what positions is the energy entirely kinetic and potential in SHM Support your answer with appropriate figure?

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• Answer:

                  (i) Simple harmonic motion. A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position. If the displacement of the oscillating particle from the mean position is small, then \[\text{Restoring}\,\text{force}\propto \,\text{Displacement}\] or \[F\propto x\] or \[F=-k\,x\] where k is positive constant called force constant or spring factor and is defined as the restoring force produced per unit displacement. The negative sign shows that the restoring force always acts in the opposite direction of displacement\[x\]. The above equation defines SHM. (ii) The energy is entirely kinetic at mean position i.e., at \[y=0\]. The energy is entirely potential at extreme positions, i.e., \[y=\pm A.\]  (iii) Total distance travelled in time period T \[=2A+2A=4A.\]
  

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• Answer:

                  When the lift is stationary \[T=2\pi \sqrt{\frac{l}{g}}\]  (ii) When the lift accelerates upwards with an acceleration of\[\text{4}.\text{5 m}/{{\text{s}}^{\text{2}}}\], \[T'=2\pi \sqrt{\frac{l}{g+a}}=2\pi \sqrt{\frac{l}{g+4.5}}\] Clearly, the time period decreases when the lift accelerates upwards.
  

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• Answer:

                  The force which tends to bring a vibrating body from its displaced position to the equilibrium position is called restoring force. When the bob of a simple pendulum is displaced through an \[\theta \] from the vertical, a restoring force equal to \[mg\] \[\sin \theta \] due to gravity acts on it.
  

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• Answer:

                  The general equation for S.H.M. is \[y=A\sin (\omega t+{{\phi }_{0}})\] As the displacement is half of the amplitude \[(y=A/2),\]so \[A/2=A\sin (\omega t+{{\phi }_{0}})\] or  \[\sin (\omega t+{{\phi }_{0}})=\frac{1}{2}\] \[\therefore \]  \[\omega t+{{\phi }_{0}}={{30}^{\circ }}\] or \[{{150}^{\circ }}\] As the two particles are going in opposite directions, the phase of one is \[\text{3}0{}^\circ \] and that of the other \[\text{15}0{}^\circ \] Hence the phase difference between the two particles \[=150-30=\mathbf{12}{{\mathbf{0}}^{\mathbf{o}}}\]
  

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• Answer:

                  (i) When the lift goes up [Fig.(a)] with, uniform velocity v, tension in the string, \[T=mg\].        The value of g remains unaffected. The period T remains same as that in stationary lift i.e., \[T=2\pi \sqrt{\frac{l}{g}}\]

  
  When the lift goes up with acceleration a [Fig. (b)], the net upward force on the bob is \[T-mg=ma\]                 \[\therefore \] \[T'=m(g+a)\] The effective value of g is \[(g+a)\] and time period is \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g+a}}\] Clearly, \[{{T}_{1}}<T\]i.e., time period decreases. (iii) When lift comes down with acceleration a [Fig. (c)], the net downward force on the bob is \[mg-T'=ma\]    \[\therefore \]  \[T'=m(g-a)\] The effective value of \[g\] becomes \[(g-a)\] and time period is  \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g-a}}\]                            Clearly, \[{{T}_{2}}>T\] i.e., time period increases.
  

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• Answer:

                                          If the ice bob is of very small size, the position of its C.G. will remain same as the ice melts. Hence its time period will remain same. If the size of the ice bob is large, then \[T=2\pi \sqrt{\frac{\frac{2{{r}^{2}}}{5l}+l}{g}}\] As ice melts, the radius r and hence the time period T will decrease. The pendulum will oscillate faster.
  

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Page 7

• Answer:

                  \[\text{T = 2 }\!\!\pi\!\!\text{ }\sqrt{\frac{\text{1}}{\text{Acceleration per unit displacement}}}\] As the acceleration per unit displacement is a constant quantity, T is not affected on changing the amplitude. (ii) \[{{\upsilon }_{\max }}=\omega A\] When amplitude is doubled, maximum velocity is also doubled. (iii) \[{{a}_{\max }}={{\omega }^{2}}A\] When amplitude is doubled, the maximum acceleration is also doubled. (iv) \[E=2{{\pi }^{2}}m{{v}^{2}}{{A}^{2}}\] i.e., \[E\propto {{A}^{2}}\] When amplitude is doubled, the energy of the oscillator becomes four times.
  

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Page 8

• Answer:

                  Suspend the known mass m form the spring and note the extension \[l\] of the spring with the metre scale. If k is the force constant of the spring, then in equilibrium                 \[kl=mg\] or \[\frac{m}{k}=\frac{l}{g}\] Time period of the loaded spring, \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{l}{g}}\] So by knowing the value of extension \[l\], time period T can be determined.
  

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• Answer:

                  As the platform moves from the mean position to the upper extreme position or from upper extreme position to mean position, the acceleration of the oscillating system acts vertically downwards and hence weight of the man will decrease. On the other hand, as the platform moves from mean position to lower extreme position and then back to mean position, the acceleration acts vertically upwards. Hence weight of the man increases.
  

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Page 10

• Answer:

                  Let k be the force constant of the full spring Then frequency of oscillation of mass m will be \[{{v}_{1}}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] When the spring is cut to one-half of its length, its force constant is doubled \[(2k)\]. Frequency of oscillation of mass m will be \[{{v}_{2}}=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\] \[\therefore \]  \[{{v}_{2}}/{{v}_{1}}=\sqrt{2}\]
  

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• Answer:

                  All trigonometric functions are periodic. The sine and cosine functions can take value between ? 1 and + 1 only. So they can be used to represent a bounded motion like S.H.M. But the functions such as tangent, cotangent, secant and cosecant can take value between 0 and \[\infty \] (both positive and negative). So those functions cannot be used to represent bounded motion like S.H.M.
  

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Page 12

• Answer:

                  Clearly, \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-\alpha x\] or  \[a=-ax\]                 Time period, \[T=2\pi \sqrt{\frac{x}{a}}=2\pi \sqrt{\frac{x}{\alpha x}}=\frac{2\pi }{\sqrt{\alpha }}\]
  

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Page 13

• Answer:

                  Displacement, \[y={{\sin }^{2}}\omega t\] Velocity,     \[\upsilon =\frac{dy}{dt}2\sin \omega t\times \cos \omega t\times \omega \] \[=\omega \sin 2\omega t\] Acceleration, \[a=\frac{d\upsilon }{dt}=\omega \times \cos 2\omega t\times 2\omega \] \[=2{{\omega }^{2}}\cos 2\omega t\] As the acceleration a is not proportional to displacement y, the given function does not represent SHM. It represents a periodic motion of angular frequency \[2\omega \]. \[\therefore \]  Time period, \[T=\frac{2\pi }{\text{Angular frequency}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }.\]
  

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• Answer:

                  Time period, \[T=2\pi \sqrt{\frac{l}{g}}\] i.e., \[T\propto {{l}^{1/2}}\] The percentage increase in time period is given by                                 \[\frac{\Delta \Tau }{T}\times 100=\frac{1}{2}\frac{\Delta l}{l}\times 100\] \[=\frac{1}{2}\times 21%=\mathbf{10}\mathbf{.5%}\]
  

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