To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that
Easy
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A body executing simple harmonic motion has a maximum acceleration equal to $ 24\,metres/se{c^2} $ and maximum velocity equal to $ 16\;metres/sec $. The amplitude of the simple harmonic motion isView Solution
- 2Equations ${y_1} = A\sin \omega t$ and ${y_2} = \frac{A}{2}\sin \omega t + \frac{A}{2}\cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions isView Solution
- 3Four simple harmonic vibrations:View Solution
${y_1} = 8\,\cos\, \omega t;\,{y_2} = 4\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right)$ ;
${y_3} = 2\cos \,\left( {\omega t + \pi } \right);\,{y_4} = \,\cos \,\left( {\omega t + \frac{{3\pi }}{2}} \right)$ ,
are superposed on each other. The resulting amplitude and phase are respectively;
- 4A particle executes simple harmonic motion and is located at $x = a, b$ and $c$ at times $t_0, 2t_0$ and $3t_0$ respectively. The frequency of the oscillation isView Solution
- 5A particle executes $S.H.M.,$ the graph of velocity as a function of displacement is :-View Solution
- 6The maximum acceleration of a particle in $SHM$ is made two times keeping the maximum speed to be constant. It is possible whenView Solution
- 7A particle executes simple harmonic motion (amplitude $= A$) between $x = - A$ and $x = + A$. The time taken for it to go from $0$ to $A/2$ is ${T_1}$ and to go from $A/2$ to $A$ is ${T_2}$. ThenView Solution
- 8A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , thereforeView Solution
- 9View SolutionAn assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is :-
- 10A mass $m = 8\,kg$ is attahced to a spring as shown in figure and held in position so that the spring remains unstretched. The spring constant is $200\,N/m$. The mass $m$ is then released and begins to undergo small oscillations. The maximum velocity of the mass will be ..... $m/s$ $(g = 10\,m/s^2)$View Solution
