Class 11 Oscillations - Oscillations Due to Spring

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 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: 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: 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}\$

Q: 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 \$\displaystyle\frac{5T}{3}\$. Then the ratio of \$\displaystyle\frac{m}{M}\$ is

From formula 1, we can say\$T=2\pi \sqrt{\dfrac{M}{k}}\$ ...(1)Also,\$\dfrac{5T}{3}=2\pi \sqrt{\dfrac{M+m}{k}}\$ ...(2)Dividing eqn 2 and 1, we get\$\dfrac{5}{3}= \sqrt{\dfrac{M+m}{M}}\$\$\therefore \dfrac{25}{9}= \dfrac{M+m}{M}\$\$\therefore \dfrac{25}{9}=1+ \dfrac{m}{M}\$Hence, answer=16/9, i.e. option B

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Oscillations Due to Spring

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Oscillations Due to a Spring

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Problems on SHM of Springs

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Hooke's Law

7 mins

Relation Between Spring Constant and Length of the Spring

8 mins

Problems Based on SHM with Spring Force - I

10 mins

Problems based on SHM with Spring Force - II

13 mins

Problems Based on SHM with Spring Force - III

12 mins

Problems Based on SHM with Spring Force - IV

9 mins

Problems Based on SHM with Spring Force - V

8 mins

Quick Summary With Stories

Oscillations Due to a Spring

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Problems on SHM of springs

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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 $\frac{5 T}{3}$ . Then the ratio of $\frac{m}{M}$ is

Medium

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Revise with Concepts

Oscillations Due to a Spring

Problems on SHM of Springs

Quick Summary With Stories

Oscillations Due to a Spring

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 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 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

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

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 35T​. Then the ratio of Mm​ 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 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:

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

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

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 $\frac{5 T}{3}$ . Then the ratio of $\frac{m}{M}$ is