Advertisements
Advertisements
प्रश्न
The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
Advertisements
उत्तर
As the block of mass M is pulled, a net resultant force is exerted by the three springs opposing the motion of the block.
Now, springs k2 and k3 are in connected as a series combination.
Let k4 be the equivalent spring constant.
\[\therefore \frac{1}{k_4} = \frac{1}{k_2} + \frac{1}{k_3} = \frac{k_2 + k_3}{k_2 k_3}\]
\[ k_4 = \frac{k_2 k_3}{k_2 + k_3}\]
k4 and k1 form a parallel combination of springs. Hence, equivalent spring constant k = k1 + k4.
\[= \frac{k_2 k_3}{k_2 + k_3} + k_1 \]
\[ = \frac{k_2 k_3 + k_1 k_2 + k_1 k_3}{k_2 + k_3}\]
\[ \therefore \text { Time peiod }, T = 2\pi\sqrt{\frac{M}{k}}\]
\[ = 2\pi\sqrt{\frac{M \left( k_2 + k_3 \right)}{k_2 k_3 + k_1 k_2 + k_1 k_3}}\]
(b) Frequency \[\left( v \right)\] is given by,
\[v = \frac{1}{T}\]
\[= \frac{1}{2\pi}\sqrt{\frac{k_2 k_3 + k_1 k_2 + k_1 k_3}{M\left( k_2 + k_3 \right)}}\]
(c) Amplitude ( x ) is given by,
\[x = \frac{F}{k} = \frac{F\left( k_2 + k_3 \right)}{k_1 k_2 + k_2 k_3 + k_1 k_3}\]
APPEARS IN
संबंधित प्रश्न
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?
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.
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?
A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.
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 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.
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.
Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.
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 the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______.
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 ______.
Motion of an oscillating liquid column in a U-tube is ______.
Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.
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 ______.
