The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is :
\[ x= A sin\left( \omega t \right) \]
x= (1cm) sin (2π/8t)= (1 cm ) sin ( π/4t)
v=dx/dt= π/4 cos( π/4t)
a= dv/dt = -(π/4)²sin (π/4t)
a= -(π/4)²sin (π/4×4/3)= -√3/32π² cm/s²
The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is :
One end of an ideal spring is fixed with a wall and the other end is fixed with a block of mass 1 kg. Force constant of spring is 100 N/m and block is performing S.H.M. with amplitude 3 cm. When the block is at left extreme position, another block of mass 3 kg moving directly towards 1 kg block with velocity 80/3 cm/s collides and gets stuck to it. The amplitude of oscillation of the combined body is :
A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. μs is the coefficient of friction between P and Q. The blocks move together performing simple harmonic motion with amplitude A. The
maximum value of the friction force between P and Q is :
For a particle executing simple harmonic motion, the displacement x is given by x = A cosωt. Identify the graph which represents the variation of potential energy (PE) as function of time t and displacement x
\[ P.E = \frac{1}{2}Kx^{2} = \frac{1}{2}KA^{2}cos^{2}\left( \omega t \right) \]
Graph I represents graph of cos²ωt.
\[ P.E = \frac{1}{2}Kx^{2} \]
Graph III represents a parabolic function
A body is executing simple harmonic motion. At a displacement x, its potential energy is E1 and at a displacement y, its potential energy is E2. The potential energy E at a displacement (x + y) is :
\[ E_{1}= \frac{1}{2}Kx^{2} \]
\[ E_{2}= \frac{1}{2}Ky^{2} \]
\[ E= \frac{1}{2}K(x+y)^{2}= \frac{1}{2}Kx^{2} + \frac{1}{2}Ky^{2} + Kxy \]
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is :
1. Initial Time Period:
\[
T = 2\pi \sqrt{\frac{M}{k}}
\]
2. New Time Period:
\[
\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}
\]
3. Divide the New Time Period by the Initial Time Period:
\[
\frac{\frac{5T}{3}}{T} = \sqrt{\frac{M + m}{M}}
\]
\[
\frac{5}{3} = \sqrt{\frac{M + m}{M}}
\]
4. Square Both Sides:
\[
\frac{25}{9} = \frac{M + m}{M}
\]
5. Solve for \( \frac{m}{M} \):
\[
\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}
\]
Answer:
\[
\frac{m}{M} = \frac{16}{9}
\]
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm–¹. The block is pulled to a distance of x = 10 cm from its equilibrium position at x = 0 cm on a frictionless surface from rest at t = 0. The kinetic energy of the block when it is 5 cm away from the mean position is
\[ K.E= \frac{1}{2}K\left( A^{2}-x^{2} \right)\]
\[ K.E= \frac{1}{2}\times 50\left( 10^{2}-5^{2} \right)/10^{4}= 0.19 J \]
The total mechanical energy of a spring-mass system in simple harmonic motion is E=1/2mω²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will :
The total mechanical energy of a particle executing simple harmonic motion is E. When the displacement is half the amplitude its kinetic energy will be :
\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]
K.E= 3E/4