12. A person stands on a bathroom-type scale, which rests on a platform suspended by a large spring. The whole system exe- cutes simple harmonic motion in a vertical direction. Describe the variation in scale reading during a period of motion.
Answers
Answer:
don't
Explanation:
don't know if I will be back until the mahal in my life that
Answer:
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kx
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/m
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,W = m(g+a)
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,W = m(g+a)= m(g+kx/m) (Since a = kx/m)
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,W = m(g+a)= m(g+kx/m) (Since a = kx/m)= mg+kx
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,W = m(g+a)= m(g+kx/m) (Since a = kx/m)= mg+kxAs particle moves downward, then acceleration will be upward.
The particle goes up from the equilibrium position. So the acceleration will be downward and increasing linearly.So, net weight of the body will be m(g+a). Here, m is the mass of the body, g is the acceleration due to gravity and a is the acceleration of the system.As the spring force kx balances the force acting on the system ma, therefore acceleration of the system will be,ma = kxSo, a = kx/mThus the net weight as particle goes up from the equilibrium position,W = m(g+a)= m(g+kx/m) (Since a = kx/m)= mg+kxAs particle moves downward, then acceleration will be upward.So weight will be,
Explanation: