Use app

>>Class 11
>>Physics
>>Oscillations
>>Oscillations Due to Spring
>> $Problems Based on SHM with Spring Force ...$

Problems Based on SHM with Spring Force - V

Q: A spring having with a spring constant 1200 N m\$\displaystyle ^{-1}\$ is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Spring constant, \$k=1200\,m^{-1}\$Mass, \$m=3kg\$Displacement, \$A=2.0\,cm=0.02m\$(i) Frequency of oscillation v, is given by the relation:\$\displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\$where, T is time period\$\therefore \displaystyle v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz\$Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation:\$a=\omega^2\,A\$where,\$\omega =\$ Angular frequency \$=\sqrt{\frac{k}{m}}\$\$A\$ = maximum displacement\$\displaystyle \therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8\, ms^{-2}\$Hence, the maximum acceleration of the mass is \$8.0\,m/s^2\$(iii) Maximum velocity, \$v_{max}=A\omega\$\$\displaystyle =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4\,m/s\$Hence, the maximum velocity of the mass is 0.4 m/s.

Q: Two masses \$m_1\$ and \$m_2\$ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, \$m_1\$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of \$m_2\$.

If only mass \${ m }_{ 2 }\$ is present the extension of spring \$l\$ is given by \${ m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k } \$, where \$k\$=Spring constantNow with \${ m }_{ 1 }\$ and \${ m }_{ 2 }\$ the extension be \$l'\$ \$({ m }_{ 1 }+{ m }_{ 2 })g=kl'\$\$l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } \$When \${ m }_{ 1 }\$ is removed, \$(l'-1)\$ acts as amplitude of oscillation\$A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k } \$\$\Rightarrow A=\cfrac { { m }_{ 1 }g }{ k } \$Now, if \${ m }_{ 2 }\$ is removed the angular frequency of oscillation \$\omega =\sqrt { \cfrac { k }{ { m }_{ 2 } } } \$(This is same as the case of \${ m }_{ 2 }\$ attach to spring)\$*\$N.B: When \${ m }_{ 1 }\$ is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass \${ m }_{ 2 }\$

Q: The frequency of vibration \$'f'\$ of a mass \$'m'\$ suspended from a spring of spring constant \$K\$ is given by a relation of this type, \$f = Cm^{x}K^{y}\$; where \$C\$ is a dimensionless quantity. The value of \$x\$ and \$y\$ are

The frequency of oscillation spring mass system \$=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ m } } \$Where, \$k\$=spring constantand \$m\$=mass with springAnd given frequency, \$f=c{ m }^{ x }{ k }^{ y }\$Comparing above two frequency we get \$x=-\cfrac { 1 }{ 2 } ,y=\cfrac { 1 }{ 2 } \$

Q: A spring balance has a scale that reads from \$0\$ to \$50 kg\$. The length of the scale is \$20 cm\$. A body suspended from this balance, when displaced and released, oscillates with period of \$0.6 s\$. What is the weight of the body ?

Maximum mass that the scale can read, \$M=50kg\$Maximum displacement of the spring = Length of the scale, \$l=20cm = 0.2 m\$Time period, \$T=0.6s\$Maximum force exerted on the spring, \$F=Mg\$where,\$g=\$ acceleration due to gravity \$=9.8\,m/s^2\$\$F=50\times 9.8=490\$\$\therefore\$ Spring constant, \$k=F/l=490/0.2=2450\,Nm^{-1}\$Mass m, is suspended from the balance.Times period, \$t=2\pi\sqrt{\dfrac{m}{k}}\$\$\therefore m=\left(\dfrac{T}{2\pi}\right)^2\times k=\left(\dfrac{0.6}{2\times 3.14}\right)^2\times 2450=22.36kg\$\$\therefore\$ Weight of the body \$=mg=22.36\times 9.8=219.16\,N\$Hence, the weight of the body is about 219 N.

Q: Two identical springs of constant are connected in series and parallel as shown in figure. A mass \$m\$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

We know that, \$n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\$In series, \$k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}\$In parallel, \$k=k_1+k_2=2k\$\$\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}\$and \$n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}\$hence, \$n_1:n_2=1:2\$

Q: Two blocks \$A\$ and \$B\$ of masses \$2m\$ and \$m\$, respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of \$A\$ and \$B\$ immediately after the string is cut, are respectively:

Two blocks A and B of mass \$2m\$and \$m\$ respectively and they are connected by a mass less and in extensible string.The equilibrium condition FBD of \$2m\$\$Kx=T+2mg\$\$=mg+2mg=3mg\$After the string is cut\$T=0\$FBD of \$2m\$\$a_2=\cfrac{3mg-2mg}{2m}\\ \quad=\cfrac{g}{2}\$FBD of \$m\$Hence \$\cfrac{g}{2}\$ and \$g\$

Q: A spring of force constant \$K\$ is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

Force constant, \$ K \propto \dfrac{1}{l}=\dfrac{c}{l}\$Let the spring cut into pieces of length \$x, 2x \$So, \$x+2x=l \Rightarrow x=\dfrac{l}{3}\$Force constant of long piece is: \$ k'=\dfrac{c}{2x}=\dfrac{Kl}{2l/3}=\dfrac{3K}{2}\$

Q: A spring of spring constant 5 x \$10 ^ { 3 }\$ N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

Work done is change in potential energyThe spring is initially stretch to 5 cm and again further stretched by 5 cm and further stretched by 5 cm\$\begin{array}{l} w={ p_{ 10 } }-{ p_{ 5 } } \\ =\frac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left[ { I{ { \left( { \frac { { 10 } }{ { 100 } } } \right) }^{ 2 } }-{ { \left( { \dfrac { 5 }{ { 100 } } } \right) }^{ 2 } } } \right] \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\left( { \dfrac { { 10 } }{ { 100 } } +\dfrac { 5 }{ { 100 } } } \right) \left( { \dfrac { { 10 } }{ { 100 } } -\dfrac { 5 }{ { 100 } } } \right) N-m \\ =\dfrac { 1 }{ 2 } \times 5\times { 10^{ 3 } }\times \dfrac { { 15 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } =2.5\times 15\times 5\times { 10^{ 3 } }\times { 10^{ -4 } } \\ =75\times 2.5\times { 10^{ -1 } }=7.5\times 2.5N-m \\ =18.75\, \, N-m \end{array}\$Option A.

Q: The time period of mass suspended from a spring is \$T\$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

\$T = 2\pi \sqrt {\dfrac{m}{k}}\$When the spring is cut into four equal parts, the spring constant of one part will becomes 4k, therefore the new time period will become\$T' = 2\pi \sqrt {\dfrac{m}{k'}}\$\$T = 2\pi \sqrt {\dfrac{m}{4k}}\$ \$T'=\dfrac{T}{2}\$

9 MinsPhysics

Language

Rate

NEXT VIDEOS

Simple Pendulum

11 mins

Working of a simple pendulum

11 mins

Time Period of Simple Pendulum

6 mins

Introduction to free and Damped Oscillations

14 mins

Introduction to Forced Oscillations

13 mins

REVISE WITH CONCEPTS

Problems on SHM of Springs

ExampleDefinitionsFormulaes

QUICK SUMMARY WITH STORIES

Problems on SHM of springs

2 mins

Important Questions

A spring having with a spring constant 1200 N m $^{- 1}$ is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Hard

View solution

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_{2}$ .

Medium

View solution

The frequency of vibration $^{'} f^{'}$ of a mass $^{'} m^{'}$ suspended from a spring of spring constant $K$ is given by a relation of this type, $f = C m^{x} K^{y}$ ; where $C$ is a dimensionless quantity. The value of $x$ and $y$ are

Medium

View solution

A spring balance has a scale that reads from $0$ to $50 k g$ . The length of the scale is $20 c m$ . A body suspended from this balance, when displaced and released, oscillates with period of $0.6 s$ . What is the weight of the body ?

Medium

View solution

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

Medium

View solution

Two blocks $A$ and $B$ of masses $2 m$ and $m$ , respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of $A$ and $B$ immediately after the string is cut, are respectively:

Medium

View solution

A spring of force constant $K$ is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

Medium

View solution

A spring of spring constant 5 x $1 0^{3}$ N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

Medium

View solution

The time period of mass suspended from a spring is $T$ . If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

Hard

View solution

Two masses of $1 k g$ and $2 k g$ respectively are connected by a massless spring as shown in figure. A force of $20 N$ acts on the $2 k g$ mass. At the instant when the $1 k g$ mass has an acceleration of $10 m s^{- 2}$ towards right, the acceleration of the $2 k g$ mass is :

Hard

View solution

Problems Based on SHM with Spring Force - V

Language

Rate

REVISE WITH CONCEPTS

Problems on SHM of Springs

QUICK SUMMARY WITH STORIES

Problems on SHM of springs

Two masses m1​ and m2​ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, m1​ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of m2​.

The frequency of vibration ′f′ of a mass ′m′ suspended from a spring of spring constant K is given by a relation of this type, f=CmxKy; where C is a dimensionless quantity. The value of x and y are

A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with period of 0.6s. What is the weight of the body ?

Two identical springs of constant are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be

Two blocks A and B of masses 2m and m, respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of A and B immediately after the string is cut, are respectively:

A spring of force constant K is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

A spring of spring constant 5 x 103 N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

Two masses of 1kg and 2kg respectively are connected by a massless spring as shown in figure. A force of 20N acts on the 2kg mass. At the instant when the 1kg mass has an acceleration of 10ms−2 towards right, the acceleration of the 2kg mass is :

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_{2}$ .

The frequency of vibration $^{'} f^{'}$ of a mass $^{'} m^{'}$ suspended from a spring of spring constant $K$ is given by a relation of this type, $f = C m^{x} K^{y}$ ; where $C$ is a dimensionless quantity. The value of $x$ and $y$ are

A spring balance has a scale that reads from $0$ to $50 k g$ . The length of the scale is $20 c m$ . A body suspended from this balance, when displaced and released, oscillates with period of $0.6 s$ . What is the weight of the body ?

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

Two blocks $A$ and $B$ of masses $2 m$ and $m$ , respectively are connected by a massless and inextensible string. The whole system is syspended by a massless spring as shown in the figure. The magnitude of acceleration of $A$ and $B$ immediately after the string is cut, are respectively:

A spring of force constant $K$ is cut into two pieces such that the one piece is double the length of the other. Then the long piece will have a force constant of:

A spring of spring constant 5 x $1 0^{3}$ N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

The time period of mass suspended from a spring is $T$ . If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

Two masses of $1 k g$ and $2 k g$ respectively are connected by a massless spring as shown in figure. A force of $20 N$ acts on the $2 k g$ mass. At the instant when the $1 k g$ mass has an acceleration of $10 m s^{- 2}$ towards right, the acceleration of the $2 k g$ mass is :