Simple Harmonic Motion

Q: A particle is executing SHM according to the equation x = Acos ωt . Average speed of the particle during the interval 0 ≤ t ≤ π/6ω

(A) $\frac{\sqrt{3}\omega A}{2}$

(B) $\frac{\sqrt{3}\omega A}{4}$

(C) $\frac{3 \omega A}{\pi}$

(D) $\frac{3 \omega A}{\pi}(2-\sqrt{3})$

Ans: (D)
Sol: Average Speed , $v_{avg} = \frac{\int_{0}^{t}\frac{dx}{dt}dt}{t}$

$v_{avg} = \frac{\int_{0}^{t}dx}{t}$

$= \frac{x(t)-x(0)}{t} $

$ = \frac{A(cos\pi/6 – 1)}{\pi/6 \omega} $

$ = \frac{3 \omega A}{\pi}(\sqrt{3} – 2 )$

since particle does not change it’s direction in the given interval , average speed

$ v_{avg} = \frac{3 \omega A}{\pi}(2-\sqrt{3})$

Q: A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is:

(A) $\frac{T}{20}$

(B) $\frac{T}{2\pi} cos^{-1}(\frac{4}{5})$

(C) $\frac{T}{2\pi} cos^{-1}(\frac{1}{5})$

(D) $\frac{T}{2\pi} sin^{-1}(\frac{1}{5})$

Ans: (B)

Solution: Particle is starting from rest, i.e. from one of its extreme position.

As particle moves a distance A/5, we can represent it on a circle as shown

$\displaystyle cos\theta = \frac{4A/5}{A} = \frac{4}{5}$

$\displaystyle \theta = cos^{-1}(\frac{4}{5}) $

$\displaystyle \omega t = cos^{-1}(\frac{4}{5}) $

$\displaystyle t = \frac{1}{\omega} cos^{-1}(\frac{4}{5}) $

$\displaystyle t = \frac{T}{2 \pi} cos^{-1}(\frac{4}{5}) $

Q: A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration √3 g m/s² .. The period of small oscillations of the pendulum about its equilibrium position is (g = π²m/s²) :

(A) 1.0 sec

(B) 2 sec

(C) 1.53 sec

(D) 1.68 sec

Ans: (A)

Solution: With respect to the cart, equilibrium position of the pendulum is shown.
If displaced by small angle θ from this position, then it will execute SHM about this equilibrium position, time period of which is given by :

$T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$

$T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$

$g_{eff} = \sqrt{g^2 + (\sqrt{3}g)^2}$

$g_{eff} = 2 g $

$T = 2\pi \sqrt{\frac{l}{g_{eff}}}$

Q: Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The
pendulum bob has mass 0.2 kg. The length of the pendulum is equal to (g = 10 m/s^2 )

(A) 2.0 m

(B) 1.8 m

(C) 1.5 m

(D) 1.2

Ans: (C)

Solution: $\frac{1}{2}m v_m^2 = 15 \times 10^{-3}$

$v_m = \sqrt{0.150} $

$ \omega A = \sqrt{0.150} m/s $

$ \sqrt{\frac{g}{L}} A = \sqrt{0.150} m/s $

L = 1.5 m

Q: A particle performing S.H.M. undergoes displacement of  A/2 (where A = amplitude of S.H.M.) in one second. At t = 0 the particle was located at either extreme position or mean position. The time period of S.H.M. can be : (consider all possible cases)

(A) 12 s

(B) 2.4 s

(C) 6 s

(D) 1.2 s

Click to See Solution :

Ans: (A),(B) ,(C) , (D)

Sol: If T be the time period ; time to go from O to Q is T/12 and from M to P is T/6 .

The displacement is A/2 when particle goes from O to Q, from O to N to Q, from O to N to O to P, and so on .

t = T/12 or t = T/4 + T/6 = 5T/12

or , t = T/2 + T/12 = 7T/12

Hence possible time period T is T = 12 sec or T = (12 × 1)/5 = 2.4 sec

or , T = (12 × 1) / 7 sec

similarly displacement is A/2 when particle goes from M to P or M to N to P .

Hence the possible time period T is

T = 1 × 6 = 6 s or T = (6 × 1)/5 s = 1.2 s

Ans. T = 1.2 s , 6 s , 2.4 s , 12 s

 

The potential energy of a particle executing SHM is given by

(a) U(x ) = k(x-a)²/2

(b) U(x)= k₁x + k₂x² + k₃ ³

(c) U(x) = A e^-bx

(d) U(x) = constant

Ans:(a )

$F -\frac{dU}{dx}$

For$ U(x)= \frac{k}{2}(x-a)^2$

$\frac{dU/}{dx} = k(x-a)$

F= -k(x-a)

This is the condition for SHM about point x= +a
Other functions do not satisfy this condition

Q: One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0) . A point particle of mass m carrying a positive charge q is attached at its other end . The entire system is kept on a smooth horizontal surface . When a point dipole p pointing towards the charge q is fixed at origin , the spring gets stretched to length l and attains a new equilibrium position . If the point mass is now displaced slightly by Δl << l from its equilibrium position and released , it is found to oscillate at frequency $\frac{1}{\delta} \sqrt{\frac{k}{m}}$ . The value of δ is ———-

Ans: (3.14)

Solution: Electrostatic force = Spring force

$F_e = F_{sp}$

$ \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3} = k l$

Now , the mass m is displaced by Δl = x from the mean position

F_net = F_sp – F_e

$F_{net} = k(l+x) – \frac{1}{4\pi \epsilon_0}\frac{2p q}{(l+x)^3} $

$F_{net} = k(l+x) – \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3(1+x/l)^3} $

$F_{net} = kl + kx – \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3}(1-\frac{3 x}{l}) $

Substituting $ \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3} = k l$

$F_{net} = kx + kl (\frac{3x}{l}) = 4 k x $ ; This force is restoring in nature .

Therefore , k_eq = 4 k

Time period , $\displaystyle T = 2\pi \sqrt{\frac{m}{4k}}$

$\displaystyle T = \pi \sqrt{\frac{m}{k}}$

Frequency , $\displaystyle \nu = \frac{1}{\pi} \sqrt{\frac{k}{m}}$

Q: An object of mass m is suspended at the end of a massless wire of length L and area of cross-section A . Young modulus of the material of wire is Y . If the mass is pulled down slightly its frequency of oscillation along the vertical direction is

(a) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{m L}{Y A}}$

(b) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y A}{m L}}$

(c) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{m A}{Y L}}$

(d) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y L}{m A}}$

Ans: (b)

Solution: Frequency of oscillation $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{k_{eq}}{m}}$

$\displaystyle Y = \frac{Stress}{Strain}$

$\displaystyle Y = \frac{F/A}{(\Delta L/L)}$

$\displaystyle F = (\frac{Y A}{L})\Delta L$

Equivalent Stiffness constant $\displaystyle k_{eq} = \frac{Y A}{L}$

$\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y A}{m L}}$

Q: When a particle of mass m is attached to a vertical spring of spring constant k and released , its motion is described by y(t) = y_o sin² ω t , Where ‘y’ is measured from the lower end of unstretched spring . The ω is

(a) $\displaystyle \frac{1}{2} \sqrt{\frac{g}{y_o}} $

(b) $\displaystyle \sqrt{\frac{g}{y_o}} $

(c) $\displaystyle \sqrt{\frac{g}{2 y_o}} $

(d) $\displaystyle \sqrt{\frac{2 g}{y_o}} $

Ans: (c)

Solution : $\displaystyle y(t) = y_o sin^2 \omega t $

$\displaystyle y(t) = \frac{y_o}{2} (1-cos2 \omega t) $

Here angular frequency of oscillation ω’ = 2 ω

$\displaystyle \omega’ = \sqrt{\frac{k}{m}} $

$\displaystyle 2 \omega = \sqrt{\frac{k}{m}} $

$\displaystyle \omega = \frac{1}{2} \sqrt{\frac{k}{m}} $

Also $\large mg = k (Amplitude ) = k \frac{y_o}{2}$

$\displaystyle \frac{k}{m} = \frac{2 g}{y_o}$

$\displaystyle \omega = \sqrt{\frac{g}{2 y_o}} $

Q: A body of mass m is situated in a potential field U(x) = U_0 (1 – cos⁡αx) when, U_0 and α are constants. Find the time period of small oscillations.