Advertisements
Advertisements
प्रश्न
A mass of 2 kg is attached to the spring of spring constant 50 Nm–1. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.
Advertisements
उत्तर
Consider the diagram of the spring block system. It is an S.H.M. with an amplitude of 5 cm about the mean position shown.
Given, the spring constant k = 50 N/m
m = mass attached = 2 kg
∴ Angular frequency `ω = sqrt(k/m)`
= `sqrt(50/2)`
= `sqrt(25)`
= 5 rad/s
Assuming the displacement function
`y(t) = Asin(ωt + phi)`
Where `phi` = initial phase
But given at t = 0, y(t) = + A
y(0) = + A = Asin(ω × 0 + `phi`)
or `sin phi` = 1 ⇒ `phi = pi/2`
∴ The desired equation is `y(t) = Asin(ωt + pi/2) = Acos ωt`
Putting A = 5 cm, ω = 5 rad/s
We get, y(t) = 5sin5t
Where t is in second and y is in centimetre.
APPEARS IN
संबंधित प्रश्न
The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find the position(s) of the particle when the speed is 8 cm/s.
A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s−1)t + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at t = 0.
The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?
A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium find the potential energy stored in the spring.
The block of mass m1 shown in figure is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m1 + m2)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
Discuss in detail the energy in simple harmonic motion.
Show that for a particle executing simple harmonic motion.
- the average value of kinetic energy is equal to the average value of potential energy.
- average potential energy = average kinetic energy = `1/2` (total energy)
Hint: average kinetic energy = <kinetic energy> = `1/"T" int_0^"T" ("Kinetic energy") "dt"` and
average potential energy = <potential energy> = `1/"T" int_0^"T" ("Potential energy") "dt"`
When a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle?
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.
A body is performing S.H.M. Then its ______.
- average total energy per cycle is equal to its maximum kinetic energy.
- average kinetic energy per cycle is equal to half of its maximum kinetic energy.
- mean velocity over a complete cycle is equal to `2/π` times of its π maximum velocity.
- root mean square velocity is times of its maximum velocity `1/sqrt(2)`.
Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.
A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = A sin`(pit)/90`. The ratio of kinetic to the potential energy of this particle at t = 210s will be ______.
The total energy of a particle, executing simple harmonic motion is ______.
where x is the displacement from the mean position, hence total energy is independent of x.
