Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m1 + m2) g sin θ
At equilibrium,
kx = (m1 + m2) g sin θ
\[\Rightarrow x = \frac{\left( m_1 + m_2 \right) g \sin \theta}{k}\]
(b) It is given that:
Distance at which the spring is pushed,
\[x_1 = \frac{2}{k}\left( m_1 + m_2 \right)g \sin \theta\]
As the system is released, it executes S.H.M.
where \[\omega = \sqrt{\frac{k}{m_1 + m_2}}\]
When the blocks lose contact, P becomes zero. (P is the force exerted by mass m1 on mass m2)
\[\therefore m_2 g \sin \theta = m_2 x_2 \omega^2 = m_2 x_2 \times \frac{k}{m_1 + m_2}\] \[ \Rightarrow x_2 = \frac{\left( m_1 + m_2 \right) g \sin \theta}{k}\]
Therefore, the blocks lose contact with each other when the spring attains its natural length.
(c) Let v be the common speed attained by both the blocks.
\[\text { Total compression } = x_1 + x_2 \]
\[\frac{1}{2} \left( m_1 + m_2 \right) v^2 - 0 = \frac{1}{2}k \left( x_1 + x_2 \right)^2 - \left( m_1 + m_2 \right) g \sin \theta \left( x + x_1 \right) \]
\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2 = \frac{1}{2}k\left( \frac{3}{k} \right) \left( m_1 + m_2 \right) g \sin \theta - \left( m_1 + m_2 \right) g \sin \theta \left( x_1 + x_2 \right)\]
\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2 = \frac{1}{2} \left( m_1 + m_2 \right) g \sin \theta \times \left( \frac{3}{k} \right) \left( m_1 + m_2 \right) g sin \theta\]
\[ \Rightarrow v = \sqrt{\left\{ \frac{3}{k} \left( m_1 + m_2 \right) \right\}}g \sin \theta\]
APPEARS IN
संबंधित प्रश्न
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
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?
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.
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 ?
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.
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.
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 rectangle plate of sides a and b is suspended from a ceiling by two parallel string of length L each in Figure . The separation between the string is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.
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.
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?
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 ______.
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.
Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.
Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.
