In an engine the piston undergoes vertical simple harmonic motion with amplitude $$7\ cm$$. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
No contact $$\Rightarrow$$ N = 0
$$\Rightarrow a_{max}=g=w^2A$$
$$w=\sqrt{g/A}$$
$$=\sqrt{\frac{10}{0.07}}$$
$$f=\frac{w}{2\pi}=\sqrt{\frac{1000}{7}}\frac{1}{2\pi}=1.9$$
An oscillator of mass $$M$$ is at rest in its equilibrium position in a potential $$V = \dfrac {1}{2} k (x - X)^{2}$$. A particle of mass $$m$$ comes from right with speed $$u$$ and collides completely inelastically with $$M$$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after $$13$$ collisions is: $$(M = 10, m = 5, u = 1, k = 1)$$.
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion 10 $$S^{-1}$$ . At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is $$ \frac {\pi} {4}$$ .
A particle is executing a simple harmonic motion. Its maximum acceleration is $$\alpha$$ and maximum velocity is $$\beta$$. Then, its time period of vibration will be:
An object of mass '$$m$$' is kept on a horizontal platform. The platform is oscillating vertically with amplitude '$$a$$' and period '$$T$$'. The weight of object at lowest position is:
A $$5$$ kg collar is attached to a spring of spring constant $$500 Nm^{-1}$$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $$10$$cm and released. The maximum acceleration of the collar is
A particle executing simple harmonic motion with an amplitude A and angular frequency $$\omega$$. The ratio of maximum acceleration to the maximum velocity of the particle is
A particle executing SHM according to the equation $$x = 5 cos[2 \pi t + \frac{\pi}{4}]$$ in SI units. The displacement and acceleration of the particle at t = $$1.5$$s is:
A mass oscillates along the x-axis according to the law, $$x = x_0 cos (\omega t - \frac{\pi}{4})$$. If the acceleration of the particle is written as $$a = A cos (\omega t + \delta)$$, then
A particle is executing simple harmonic motion with amplitude of $$0.1 m$$. At a certain instant when its displacement is $$0.02$$ and its acceleration is $$0.5 { m }/{ { s }^{ 2 } }$$, the maximum velocity of the particle is (in $${ m }/{ s }$$) :
A particle is executing simple harmonic motion with an amplitude of 0.02 meter and frequency 50$$\mathrm { Hz }$$ The maximum acceleration of the particle is