Advertisements
Advertisements
Question
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)`.
Advertisements
Solution
a, b and d
Explanation:
In the case of S.H.M, the average total energy per cycle
= Maximum kinetic energy (K0)
= Maximum potential energy (U0)
Average KE per cycle = `(0 + K_0)/2 = K_0/2`
Let us write the equation for the SHM x = a sin ωt.
Clearly, it is a periodic motion as it involves a sine function.
Let us find velocity of the particle, `v = (dx)/(dt) = d/(dt) (a sin ωt) = aω cos ωt`
Mean velocity over a complete cycle,
`v_"mean" = (int_0^(2pi) ωa cos θd θ)/(2pi)`
= `(ωa[sin θ]_0^(2pi))/(2pi)`
= 0
So, `v_"mean" ≠ 2/pi v_"max"`
Root mean square speed,
`v_(rms) = sqrt((v_"min"^2 + v_"max"^2)/2`
= `sqrt((0 + v_"max"^2)/2`
`v_(rms) = 1/sqrt(2) v_"max"`
APPEARS IN
RELATED QUESTIONS
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?
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan−1 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for the second time ?
The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1 N/m. What mass should be attached to 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?

The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.

Repeat the previous exercise if the angle between each pair of springs is 120° initially.
Find the elastic potential energy stored in each spring shown in figure when the block is in equilibrium. Also find the time period of vertical oscillation of the block.

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?
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.

Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
What is the amplitude of oscillation?
An object of mass 0.5 kg is executing a simple Harmonic motion. Its amplitude is 5 cm and the time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = `T/4` s starting from the mean position? Assume that the initial phase of the oscillation is zero.
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 ______.
