A particle moves with simple harmonic motion is a straight line. In first τ sec, after starting from rest…

Q: A particle moves with simple harmonic motion is a straight line. In first τ sec, after starting from rest it travels a distance ‘ a ‘ and in next τ sec, it travels 2a , in same direction, then

(a)Amplitude of motion is 3a

(b)Time period of oscillations is 8π

(c)Amplitude of motion is 4a

(d)Time period of oscillations is 6π

Ans: (d)

Sol: As particle starting from rest .

x = A cosωt

At t= 0 ; x = A

When t = τ ; x = A – a

When t = 2τ ; x = A – 3a

Therefore ,

A – a = A cosωτ

A – 3a = A cos2ωτ

using formula ;

cos2ωτ = 2cos2ωτ – 1

$\large \frac{A-3a}{A} = 2 (\frac{A-a}{A})^2 – 1$

on solving this we get

A = 2a

2a – a = 2a cosωτ

cosωτ = 1/2

ωτ = π/3

$\large \frac{2\pi}{T}\tau = \frac{\pi}{3}$

A small block is connected to one end of a massless springs of unstretched length 4.9 m…

Q: A small block is connected to one end of a massless springs of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency ω = π/3 rad/s. simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point P is at a horizontal distance of 10m from O. if the pebble hits the block at t=1s, the value of v is
(take g=10 m/s2)

Q & A

(a) $ \sqrt{50}$ m/s

(b) $ \sqrt{51}$ m/s

(c) $ \sqrt{52}$ m/s

(d) $ \sqrt{53}$ m/s

Ans: (a)

Sol: time of flight t = 1 sec

$\large \frac{2v sin\theta}{g} = 1$

$\large \frac{2 v sin45^o}{10} = 1$

$v = \sqrt{50}$ m/s

A wooden block perform SHM on a frictionless surface with frequency ν0. The block carries a charge +Q on its surface…

Q: A wooden block perform SHM on a frictionless surface with frequency ν0. The block carries a charge +Q on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the SHM of the block will be

Q & A

(a)Of the same frequency and with shifted mean position

(b)Of the same frequency and with same mean position

(c)Of the changed frequency and with shifted mean position

(d)Of the changed frequency and with same mean position

Ans: (a)

Sol: As constant external force can only change the mean position . When electric field is switched on , mean position will be obtained after a compression of x0

$\large k x_0 = Q E$

$\large x_0 = \frac{Q E}{k}$

The mass M shown in the figure oscillates in simple harmonic motion with amplitude A…

Q: The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. the amplitude of the point P is

Q & A

(a) $\large \frac{k_1 A}{k_2}$

(b) $\large \frac{k_2 A}{k_1}$

(c) $\large \frac{k_1 A}{k_1 + k_2}$

(d) $\large \frac{k_2 A}{k_1 + k_2}$

Ans: (d)

Sol: $\large x_1 + x_2 = A $ …(i)

$\large k_1 x_1 = k_2 x_2$ …(ii)

on solving (i) & (ii)

$\large x_1 = \frac{k_2 A}{k_1 + k_2}$

A uniform rod of length L and mass M is provided at the centre. Its two ends are attached to two springs…

Q: A uniform rod of length L and mass M is provided at the centre. Its two ends are attached to two springs of equal spring constants k . The springs are fixed to rigid supports as shown in the figure. And rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is

Q & A

(a) $\large \frac{1}{2\pi} \sqrt{\frac{2 k}{M}}$

(b) $\large \frac{1}{2\pi} \sqrt{\frac{k}{M}}$

(c) $\large \frac{1}{2\pi} \sqrt{\frac{6 k}{M}}$

(d) $\large \frac{1}{2\pi} \sqrt{\frac{24 k}{M}}$

Ans: (c)

Sol: The rod is gently pushed through a small angle θ

$\large \theta = \frac{x}{L/2} = \frac{2 x}{L}$

Restoring Torque $\large \tau = – (2 k x)\frac{L}{2}$

$\large I \alpha = – \frac{k}{L} \frac{L \theta}{2}$

$\large \alpha = – \frac{k L^2}{2 I} \theta$

$\large \alpha = – \frac{k L^2}{2 (M L^2/12)} \theta$

$\large \alpha = – \frac{6 k}{M} \theta $

Frequency $\large f = \frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$