DC Pandey Solutions: Simple Harmonic Motion - 1

# DC Pandey Solutions: Simple Harmonic Motion - 1 | Physics Class 11 - NEET PDF Download

## Introductory Exercise 11.1

Q.1. A particle moves under the force F(x) = (x2 - 6x) N, where x is in metres. For small displacements from the origin, what is the force constant in the simple harmonic motion approximation?

Sol.

F(x)=x2 -6x
Fr = F(x+ Δx) - F(x)
= (x + Δ)2 - 6(x+ Δx) - x2 + 6x
= 2xΔx - 6Δx = -(6 - 2x) Δx
= - 6Δx for x = 0
kΔx ⇒ k = 6N/m

Q.2. At A= A/2, what fraction of the mechanical energy is potential? What fraction is kinetic? Assume potential energy to be zero at the mean position.

Sol.

Q.3. The initial position and velocity of a body moving in SHM with period T = 0.25 s are x = 5.0 cm and v = 218 cm/s. What are the amplitude and phase constant of the motion?

Sol.

= 10cm
x = A sin (ωt + θ)

Q.4. A cart of mass 2.00 kg is attached to the end of a horizontal spring with force constant k = 150 N/m. The cart is displaced 15.0 cm from its equilibrium position and released. What are (a) the amplitude (b) the period (c) the frequency (d) the mechanical energy (e) the maximum velocity of the cart? Neglect friction.

Sol.

(a) Maximum displacement
= 15 cm = Amplitude.

Q.5. A body of mass 0.10 kg is attached to a vertical massless spring with force constant 4.0 x 10" N/m. The body is displaced 10.0 cm from its equilibrium position and released. How much time elapses as the body moves from a point 8,0 cm on one side of the equilibrium position to a point 6.0 cm on the same side of the equilibrium position?

Sol.

A = 10cm
8 cm = 10 sin ωt1 and 6 cm = 10 sin ωt2

= 1.4 × 10-3s

Q.6. Does a 2.0 kg particle undergo SHM according to x = 1.5 sin  (in SI units

(a) What is the total mechanical energy of the particle?
(b) What is the shortest time required for the particle to move from x = 0.5 m to x = - 0.75 m?

Sol.

Q.7. A body of mass 200 g is in equilibrium at x = 0 under the influence of a force F(x) = (-100x + 10x2) N
(a) If the body is displaced a small distance from equilibrium, what is the period of its oscillations?
(b) If the amplitude is 4.0 cm, by how much do we error in assuming that F(x) = - kx at the end points of the motion.
Sol.

The document DC Pandey Solutions: Simple Harmonic Motion - 1 | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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## Physics Class 11

102 videos|411 docs|121 tests

## FAQs on DC Pandey Solutions: Simple Harmonic Motion - 1 - Physics Class 11 - NEET

 1. What is simple harmonic motion?
Ans. Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth about its equilibrium position under the influence of a restoring force that is directly proportional to its displacement from the equilibrium position.
 2. What are the characteristics of an object in simple harmonic motion?
Ans. The characteristics of an object in simple harmonic motion include a periodic motion, a constant amplitude, a constant frequency, and a sinusoidal displacement-time graph.
 3. How is the period of simple harmonic motion related to its frequency?
Ans. The period of simple harmonic motion is the time taken for one complete oscillation, while the frequency is the number of oscillations per unit time. The period and frequency are inversely related and can be calculated using the formula: frequency = 1 / period.
 4. What is the relationship between the amplitude and energy of an object in simple harmonic motion?
Ans. In simple harmonic motion, the amplitude is directly proportional to the maximum potential energy and the maximum kinetic energy of the object. As the amplitude increases, so does the energy of the object.
 5. How does damping affect simple harmonic motion?
Ans. Damping refers to the gradual decrease in amplitude over time in a simple harmonic motion. It is caused by external forces such as friction or air resistance. Damping reduces the energy of the system and can eventually cause the oscillations to stop.

## Physics Class 11

102 videos|411 docs|121 tests

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