QUESTIONS 1. Is the tension in the string of a simple pendulum constant throughout the oscillations? 2. The amplitude of a simple harmonic oscillator is doubled. How does this effect (i) the period  (ii)the total energy (iii) the maximum velocity of the oscillator? 3. How are time p...

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Answer--In simple pendulum, when bob is in deflection position, the tension in the string is T=mgcosθ. Since, the value of θ is differenet at different positions, hence tension in the string is not constant throughout the oscillation.

1.ans:  No the tension in the string of a simple pendulum is not constant throughout the motion                                     

v is velocity of the bob and theta is the angle made by the string with vertical axis.

as v and θ both are variable T is also variable.

T is maximum at mean position as at mean position v is max and the angle is minimum.

2.ans: If the amplitude of the simple harmonic oscillator is doubled then:                                                                                                                                                                                                                                               For time period:        Time period is independent of the amplitude.                                                                                                                                                                                                                                                                For total energy: E=  where r is the amplitude. If r=2r then then the expression becomes E'= Thus, the total energy increases by 4 times.                                For  max. velocity: V=rω , where r=amplitude. Thus, max velocity becomes V'=(2r).ω=2.rω. Max. velocity increases by 2 times.

3.ans: Time period of SHM is given by: T=In both the cases, time period doesn't show any dependency on the amplitude of the SHM. (In the derivation of the equations of motion of a harmonic oscillator we assume that the amplitude is small — so even after doubling it, we still remain well within this approximation.)  From logical point of view it can be observed as  (since the pendulum or the mass of the spring have to travel a larger distance when oscillating with a larger amplitude. But the force acting on the mass is larger, overall, as well, and so is the maximum velocity. The mass has to travel a greater distance, but is faster, perfectly compensating for it.). Thus, doubling the amplitude has no effect on its time period and its inverse i.e. frequency.

4.ans:  The time period of a simple pendulum is given by:   The body will be in a state of weightlessness inside a satellite so that the effective value for g for it is zero. If g=0, T is equal to infinity. Hence, the pendulum does not oscillate inside the satellite and the simple pendulum experiment cannot be performed inside a satellite.

5.ans: The first condition is not sufficient as "acceleration∝  displacement" doesn't specify the direction of acceleration as acceleration in SHM is always direction in the opposite direction of that of displacement.

6.ans:  Consider a particle in SHM whose diagrammatic representation can be done in a circle. Then, one complete oscillation(distance) in SHM is represented as the circumference of the circle and its angular displacement is represented by angle travelled in it with angular velocity  'ω'.

 Then, total angular distance travelled by it in one time period becomes 2π as it is a circle.  T==. Hence, proved.